
LIBRARY OF CONGRESS. 

Cli\|o5j5lpjright No 



Shell. j6f__2. 7 



UNITED STATES OF AMERICA. 



J 




.y 



ELEMENTS 



OF 



Analytic Geometry 



BY 



JOSEPH JOHNSTON HARDY 

Professor of Mathematics and Astronomy in I^afayette College 



r 



6'' 



EASTON, PA. : 
CHEMICAL PUBLISHING COMPAN 

1S97. 

(^All rights reserved.) 




ty*5 \ ~ 

TWO G^i^^'^^ v^rf-nvrv^^ 



Copyright, 1897, by Edward Hart. 



PRKKACB 



This book has been written not to expound anything new 
but to provide a convenient and useful class manual. It 
seeks to give first discipline and second useful knowledge. 

This discipline should consist first in training to think, and 
second in training to express thought. 

This training to think will consist mainly in training to 
form clear, precise, and correct mathematical concep- 
tions and to form trains of deductive reasoning constructed 
with perfect logical soundness and cogency. Hence the 
student should be frequently required to give clear, precise, 
and correct definitions of the terms he is using, to show every 
point, line and angle discussed in a neat clear and correct 
figure and to translate his final equation into ordinary lan- 
guage until at last each letter and symbol in it shall at once, 
and clearly, express to him the magnitude or the operation for 
which it stands. He should also be required to state clearly 
and correctly each step in the argument, to state it in its log- 
ical order and to fortify each step by quoting the proof for 
it ; so that when he reaches his final equation he shall feel 
that it must necessarily be true. Hence an attempt has been 
made in this book to cast the demonstrations as lar as possible 
in the form of those given in elementary geometry. 

In explaining his work at the blackboard, the student 
should be required first to state his theorem in the best 
language he can command, then to illustrate its meaning by 
applying it to the figure, then to give his demonstration with 
all logical rigor and clearness of language, and finally to 
draw his conclusion correctly ; doing all this in the spirit of 



iv PREFACE 

one who feels that he has something important which he 
wishes to persuade his hearers is true. 

Originality and skill in the art of manipulating the sym- 
bols and in applying the methods of Analytic Geometry 
will come from the solution of a large number of well-selected 
examples. 

In the making of this manual, Mr. James G. Hardy has 

helped me very much by judicious and scholarly criticism and 

suggestion. 

Joseph J. Hardy. 



T^ABI^K OF CONXENTPS 



CHAPTER I 

PAGE. 

Constants and Variables 

A constant ^ 

A variable ■> 

The change ^ 

CHAPTER II 

Location of Points in a Plane 

Coordinate axes 4 

Coordinates of a point 4 

Method of locating a point in a plane 5 

CHAPTER III 

Construction of Loci 

A locus 14 

The equation of a locus 16 

The locus oi y = sin ;r 17 

The locus oi y = tan ;r 18 

The locus of jv = sec ^ 19 

CHAPTER IV 

The Intersection of Loci 

Rule for finding where a locus cuts the X axis 22 

Rule for finding where a locus cuts the Y axis 23 

Rule for finding where loci cuts each other 25 

CHAPTER V 

The Straight Line 

The intercept » 26 

The inclination of a line 26 

The slope 26 

The equation of a straight line in terms of its slope and Y intercept 26 

The equation of a straight line through one fixed point 30 

The equation of a straight line through two fixed points 33 

The length of the line joining two fixed points 35 

The angle between two straight lines meeting at a point 36 

Condition for perpendicularity is i -|-\y'5 ^ o 37 

The equation of a straight line in terms of both intercepts 39 

The equation of a straight line in terms of the perpendicular on it 

from the origin and the inclination of this perpendicular- . • 40 



vi TABLE OF CONTENTS 

The equation of a straight line in terms of the perpendicular on it 
from the origin and the angles which this perpendicular 
makes with both axes 42 

Every equation of the first degree containing two variables only is 

the equation of a straight line 43 

Given the equation of a straight line in the form Ax -\- By -f- C= o 

to find its equation in the form x cos a -\-y sin a — p z= o •' 44 

The distance from the point x' , y' to the line x cos a -\- y sin a — p 

= 47 

CHAPTER VI 

Oblique Axes 
The equation of a straight line in terms of its slope and Y inter- 
cept 51 

The equation of a straight line through one fixed point 53 

The equation of a straight line through two fixed points 55 

CHAPTER VII 

Transfer motion of Coordinotes 

To transform to a system of parallel axes having a new origin.. • • 58 

To transform a new system of axes having the same origin 59 

To transform from any pair of rectilinear axes to any other pair . . 64 

CHAPTER VIII 

Polar Coordinotes 

The initial line 67 

The pole * 67 

The radius vector . ". 67 

The vectorial angle 67 

The vectorial arc 67 

The polar equation of a straight line 69 

CHAPTER IX 

The Ellipse 

Definition of the ellipse • 72 

Foci 73 

Focal radii 73 

Vertices 73 

Transverse axis 73 

Center 73 

Conjugate axes 73 

The sum of the focal radii to any point on an ellipse 73 

The foci are equally distant from the center 74 

The equation of an ellipse referred to its axes 74 

The semi-transverse axis is equal to its distance from the focus to 

the extremity of the conjugate axis 76 

The circle is an ellipse 77 



TABLE OF CONTENTS vii 

The equation of a circle, the center being at the origin 77 

The equation of the circle when its center is not at the origin 78 

The circle circumscribed about an ellipse 79 

Corresponding ordinates > 79 

The relation between the ordinates of an ellipse and the circum- 
scribed circle 79 

The relation between the ordinates of an ellipse and those of the 

inscribed circle 80 

To construct an ellipse the axes being given 81 

The relation between the ordinates of any two points on an ellipse 82 

The parameter or latus rectutn 84 

The ellipse is symmetrical with respect to both axes 85 

The eccentricity 85 

The length of the focal radius 86 

Equation of the tangent to the ellipse 88 

The slope of the tangent 90 

Equation of the tangent to the circle 90 

The subtangent 91 

To draw a tangent to an ellipse from a point on it 92 

The normal 92 

Equation of the normal 92 

The subnormal 94 

Equation of the normal to the circle 95 

The normal bisects the interior and the tangent the exterior angle 

between the focal radii 96 

To draw a tangent to an ellipse at a given point on it 97 

A chord 97 

A complete system of parallel chords 98 

The diameter 98 

Equation of the diameter 98 

The diameter passes through the center 100 

Relation between the inclination of a diameter and that of its sys- 
tem of bisected chords loi 

If a diameter bisect a system of chords parallel to a second diame- 
ter the second diameter will bisect a system parallel to the 

first loi 

Conjugate diameters 102 

The relation between the inclination of a diameter and that of its 

conjugate 102 

The tangent to an ellipse at the extremity of any diameter 103 

The two tangents at the extremity of any diameter are parallel to 

each other 104 

Given the coordinates of the extremity of a diameter to find the 

coordinates of the extremity of its conjugate 104 

The diameter is bisected at the center 107 



viii TABLE OF CONTENTS 

The sum of the squares of any pair of conjugate diameters 107 

Area of the parallelogram circumscribed about an ellipse 109 

x\rea of an ellipse 112 

If the inclinations of two diameters are supplementary angles the 

diameters are equal 1 14 

The diagonals of the rectangle on the axes are conjugate diameters 115 
Conjugate diameters whose inclinations are supplementary angles 116 

Equi-conjugate diameters 117 

The directrix 117 

Focal radius 117 

Directral distance 117 

The ratio between the focal and directral distances of any point 

on an ellipse 118 

Distance from the center to the directrix 120 

Equation of the ellipse referred to any pair of conjugate dianfeters 120 
Equation of the tangent to an ellipse referred to any pair of con- 
jugate diameters 123 

Equation of the chord joining the points of tangency of two tan- 
gents referred to any pair of conjugate diameters 124 

The same referred to the axes of the ellipse 126 

The two tangents at the extremity of any chord 127 

The tangents at the extremity of any focal chord 128 

The locus of the intersection of two perpendicular tangents 131 

The locus of the intersection of two tangents at the extremities of 

a chord which revolves about a fixed point 134 

Supplemental chords 135 

A chord parallel to any diameter 136 

Polar equation of the ellipse, the right hand focus being the pole- 137 
Polar equation of the ellipse, the left hand focus being the pole. • 138 

Another form of the polar equation of the ellipse 138 

Polar equation of the ellipse, the center being the pole 139 

Second polar equation of the ellipse, the center being the pole 140 

CHAPTER X 

The Hyperbola 

To draw an hyperbola 143 

Corollary 144 

The foci i44 

The focal radii 144 

Vertices 144 

The transverse axis 144 

The center 144 

The conjugate axis 145 

The difference of the focal radii of any point on an hyperbola. • • • 145 
Equation of the hyperbola 146 



TABLE OF CONTENTS ix 

Any equation of an ellipse may be changed into the corresponding 

equation of the hyperbola 147 

The equilateral hyperbola 148 

The ordinates of an hyperbola and of an equilateral hyperbola 

having the same transverse axis compared 148 

Conjugate hyperbolas 150 

The equation of the conjugate to any hyperbola 150 

The squares of the ordinates of any two points on an hyperbola . . 151 

Ordinates at equal distances from the center 153 

The hyperbola is symmetrical with respect to the axes 153 

The parameter 153 

The eccentricity 154 

The values of the focal radii at any point on an hyperbola 155 

Equation of the tangent to an hyperbola 156 

The slope of the tangent 158 

The subtangent 158 

Length of the subtangent 158 

The subtangents of several hyperbolas having the same transverse 

axis 158 

The normal 159 

Equation of the normal 159 

Slope of the normal 160 

Subnormal 160 

Length of subnormal 160 

The tangent bisects the interior and the normal the exterior angle 

between the focal radii - . • . 161 

To draw a tangent to an hyperbola 163 

A chord 163 

The bisector of a system of parallel chords 163 

The diameter 163 

Equation of the diameter 163 

The diameter passes through the center 165 

The inclination of any diameter multiplied by the inclination of its 

system of bisected chords 165 

If any diameter bisect a system of chords parallel to a second 

diameter 166 

Conjugate diameters 167 

The inclination of any diameter multiplied by the inclination of 

its conjugate • • ^ 167 

The tangent at the extremity of any diameter is parallel to the 

conjugate diameter 167 

The two tangents at the extremities of any diameter are parallel • • 169 
The four tangents at the extremities of two conjugate diameters- 169 
Given the coordinates of the extremity of any diameter to find the 

coordinates of the extremities of its conjugate 169 



X TABLE OF CONTENTS. 

All diameters are bisected by the center 171 

The difference of the squares of the two conjugate diameters of an 

hyperbola 172 

The parallelogram formed by tangents to two conjugate hyperbolas 174 

The directrix ^ .... 176 

The focal distance of any point on an hyperbola 176 

The directral distance of any point on an hyperbola 176 

The ratio between the focal and directral distances of any point on 

an hyperbola 177 

The focal distance is greater than the directral distance 179 

The distance from the center to the directrix 179 

Equation of the hyperbola referred to any pair of conjugate 

diameters 180 

Equation of the tangent to an hyperbola referred to any pair of 

conjugate diameters 183 

Equation of the chord joining the points of tangency of two tan- 
gents referred to conjugate diameters 184 

Equation of the same chord referred to the axes of the hyperbola 186 

The two tangents at the extremity of any chord 187 

The two tangents at the extremities of any focal chord 188 

The locus of the intersection of two perpendicular tangents. .... 190 
The locus of the intersection of two tangents at the extremity of a 

chord which revolves about a fixed point 193 

Supplemental chords 195 

A chord parallel to any diameter of an hyperbola 195 

The polar equation of an hyperbola • 196 

CHAPTER XI 

The Asymptotes 
Equation of the asymptotes 200 

The asymptotes continually approach the hyperbola 201 

The asymptotes continually approach the conjugate hyperbola. .. 203 

Second definition of the asymptote 203 

The asymptote as the limiting case of the tangent 203 

Equations of the asymptotes referred to conjugate diameters 205 

The asymptotes are the diagonals of every parallelogram whose 

sides are parallel to two conjugate diameters 206 

Equation of the hyperbola referred to the asymptotes- ^ 208 

Equation of the tangent to the hyperbola referred to the asymp- 
totes 210 

The segment of any tangent lying between the asymptotes 212 

The product of the segments of the asymptotes between the cen- 
ter and any tangent 213 

The area of the triangle formed by any tangent and the segments 

of the asymptotes between the center and this tangent 213 



TABLE OF CONTENTS. xi 

CHAPTER XII 

The Parabolo 

The focus 216 

The directrix '• 216 

The focal radius 216 

The axis 216 

The vertex 216 

Methods of drawing a parabola 217 

Equation of the parabola 219 

The squares of ordinates 220 

Ordinates at equal distances from the vertex 220 

The principal parameter 220 

Equation of the tangent to the parabola 221 

Slope of the tangent 221 

The vertex bisects the subtangent 221 

Equation of the normal to the parabola 221 

Slope of the normal 221 

The subnormal is constant 221 

The diameter 221 

The tangent bisects the exterior and the normal the interior angle 
between the focal radius of any point and a diameter drawn 

from that point 222 

The equation of the parabola referred to any diameter and the 

tangent at its extremity 223 

Ordinates to any diameter at equal distances from its extremities 225 

Every diameter bisects a system of parallel chords 226 

The parameter of any diameter 226 

The value of the parameter of a diameter » ^ 226 

The squares of ordinates to any diameter 227 

Equation of the tangent referred to any diameter and the tangent 

at its extremity 228 

Equation of the chord which joins the points of tangency of two 

tangents 230 

The locus of intersection of two tangents at the extremities of a 

chord which revolves about a fixed point 232 

Tangents at the extremities of any focal chord meet on the direc- 
trix, etc 233 

Tangents at the extremities of any focal chord are perpendicular 

to each other 236 

Polar equations of the parabola 238 

CHAPTER XIII 

The Conic 

The focus • 240 

The directrix 240 



xii TABLE OF CONTENTS 

The ellipse is a conic 240 

The hyperbola is a conic • 240 

The parabola is a conic 241 

The polar 241 

The pole 241 

The polar of a point on any diameter of an hyperbola 242 

The polar of a point on any diameter of an ellipse 243 

The polar of a point on any diameter of a paraVjola 244 

The polar of any point on an axis 245 

The distance from the center along any diameter to the polar of 

any point on that diameter 245 

for the hyperbola 245 

ellipse 246 

The distance from the extremity of any diameter of a parabola to 

the polar of any point on that diameter 246 

The directrix is the polar of the focus for the hyperbola 247 

ellipse 248 

parabola 248 

CHAPTER XIV 

General Equation of the Second Degree 

Part first 252 

Case I. B''—AC<io 253 

Case 2. B''— ACyo 257 

Case 3. B'^ — AC=o 259 

Part second 261 

Summary • 263 

Every equation of the second degree is the equation of a conic. • • • 263 

Non-Conics 

Definition of non-conics 264 

Definition of higher plane loci 264 

Definition of the lemniscate 264 

Equation of the lemniscate in rectangular ordinates 265 

Polar equation of the lemniscate 266 

Definition of the cissoid • - 266 

Equation of the cissoid in rectangular coordinates 267 

Equation of the cissoid in polar coordinates 268 

Definition of the witch 268 

Equation of the witch in rectangular coordinates 269 

Definition of the conchoid • 270 

Equation of the conchoid in rectangular coordinates 270 

Equation of the conchoid in polar coordinates 272 

Definition of the lima^on of Pascal 272 

Equation of the lima9on in rectangular coordinates 273 

Equation of the cardioid 276 

Equation of the limagon in polar coordinates 278 



TABLE OF CONTENTS xiii 

Definition of a spiral 279 

Definition of the logarithmic spiral 279 

Equation of the logarithmic spiral 280 

To construct the logarithmic spiral whose equation is r = 26 280 

Definition of the spiral of Archimedes 281 

Equation of the spiral of Archimedes 282 

The h5'^perbolic spiral 282 

Equation of the hyperbolic spiral 283 

Definition of the parabolic spiral 284 

Equation of the parabolic spiral 285 

Definition of the lituus 285 

Equation of the lituus 286 

Definition of the logarithmic curve 286 

Equation of the logarithmic curve 286 

Definition of the cycloid 287 

Equation of the cycloid 288 

A second equation of the cycloid • • • • 290 

Other trigonometric loci 291 



SOLID ANALYTIC GEOMETRY 



CHAPTER I .. 

Points ond Directions in Space 

The position of a point in space 293 

The coordinate planes 293 

The origin 293 

The coordinate axes 293 

The axes are perpendicular to each other 294 

The coordinates of a point • 294 

The distance from any point to the origin 296 

The direction cosines of a line through the origin 296 

The coordinates of a point in terms of the direction cosines 296 

The sum of the squares of the direction cosines is i 297 

The distance between any two points 297 

The angle between two lines in space 298 

The direction angles of any line in space 298 

The direction cosines of any line in space 298 

The projections of a line on the coordinate axes 298 

Values of the direction cosines when three quantities proportional 

to them are given 299 

Direction of a line determined by three quantities proportional to 

the direction cosines 300 



xiv TABLE OF CONTENTS 

The directors of a line 300 

The direction cosines of a line are directors of that line 300 

If the directors of any line are proportional to the directors of a 

second line, the two lines are parallel 301 

The cosine of the angle between two lines 302 

Condition that two lines be perpendicular 303 

The sine of the angle between two lines 304 

CHAPTER II 

Projections 

To project any line upon another 305 

Positive and negative direction 305 

Meaning of ABx, ABy, ABz 305 

The projection of a broken line 305 

The projection of any closed contour 306 

Projection of two broken lines having the same extremities 307 

The projection of any straight line upon a second 308 

If two straight lines are parallel, their projections on the same 

plane are parallel 309 

CHAPTER III 

Tronsformation of Coordinates 

Transformation to a system of parallel axes 311 

Transformation to any other rectangular system having the same 

origin 313 

New coordinates in terms of the old table 314 

Rules for using the table of §408 - 315 

Relations between the direction cosines 315 

CHAPTER IV 

Spherical Coordinates of a Point in Space 

Definition 316 

The radius vector 316 

The latitude of a point • 316 

The longitude of a point 316 

The spherical coordinates of a point 316 

The rectangular coordinates of a point in terms of its spherical co- 
ordinates - • 317 

The direction cosines of the radius vector of a point in terms of 

the latitude and longitude 317 

CHAPTER V 
The Plane 

Equation of the plane • • 318 

Planes parallel to the coordinate planes 319 

Every equation of the first degree in three variables only is the 

equation of a plane 319 



TABLE OF CONTENTS xv 

The perpendicular from the origin to a plane 321 

Equation of a plane parallel to a given plane 321 

Distance between two planes 321 

Distance from a point to a plane x cos a -\- y cos [i -f- ^ cos y — p =0 322 

Distance from a point to a plane Ax + By -\- Cz ~\- D ^= o 322 

The traces of a plane 322 

The intercepts of a plane 324 

CHAPTER VI 
The Straight Line 

The equations of a straight line in space 325 

Second form of the equations of a straight line 327 

Third form of the equations of a straight line 329 

CHAPTER VII 

Surfaces 

A surface may be represented byy*[;r, y^ z] =o-' 334 

Definition of the equation of a surface 335 

To find where a straight line cuts a surface 335 

To find where two surfaces intersect 336 

CHAPTER VIII 
Quodrics 

The quadric 337 

Intersection of a straight line with a quadric 337 

A chord 338 

Every section of a quadric made by a plane is a conic 338 

The center 340 

A diameter 340 

Equation of a quadric referred to its center 341 

Definition of central quadrics 343 

Definition of non-central quadrics 343 

CHAPTER IX 

Central Quadrics 

Equation of a central quadric 344 

Four forms of the equation of the central quadric 346 

The ellipsoid 346 

The equation of the ellipsoid 346 

The principal axes 349 

The ellipsoid of revolution 349 

Prolate ellipsoid 350 

Oblate ellipsoid 350 

The sphere 350 

The hyperboloid of one nappe , 350 

Equation of the hyperboloid of one nappe 350 



xvi TABLE OF CONTENTS 

The hyperboloid of revolution of oue nappe 352 

The hyperboloid of two nappes 353 

Equation of the hyperboloid of two nappes 353 

The hyperboloid of revolution of two nappes 356 

CHAPTER X 
Non-Central Quodrics 

Condition that a quadric be non-central 357 

The elliptic paraboloid 357 

The equation of the elliptic paraboloid 358 

The hyperbolic paraboloid 359 

Equation of the hyperbolic paraboloid 359 

Appendix 361 



ERRATA 



Page 16, line 8, for "of a" read "of the." 

Page 18, Fig. 9, YOY' should be drawn through the left hand inter- 
section of the curve and X'X. 

Page 25, line 2, for "curves" read "loci." 

Page 113, lines 16, 17, 19, for "=" read "=." 

Page 117, line 11, for "opposite" read "same." 

Page 117, line 12, for "from" read "as." 

Page 122, line 2, for "+«2 s^^s ^/ _^^2 ^os 6*0 " read "+(^2 gj^a ^/_|_ 

Page 238, line 2, for "§ 15" read "[15]." 

Page 261, line 15, for ''riB'^ix -\ j " read "2^^ ( ^ H j." 

Page 339, lines 16, 17, for " = " read "^." 

Page 345, line 10, for " — -" read " — ^2'^." 



Page 359, line 22, for "ADEA'GH" read "ADEA'GH, Fig. 163." 



NoTiCK. — The references to geometry and trigonometry will be found 
in the appendix, page 361. 



ANALYTIC GEOMETRY 



CHAPTER I 

Constants and Voriablcs 




Fig. I 

I. In Fig. I let the point P move continually along the cir- 
cumference of the circle. Then the sine PK, the cosine PH, 
and the tangent SO will change their values continually and 
may be made to take an infinite number of different values 
in consequence of this change in the position of the point P. 

The radius CP, however, will always retain the same value 
throughout the operation of this change in the position of the 
point P. 

Hence we call the sine PK, the cosine PH, and the tangent 
SO variables, but we call the radius CP a constant. 




N P' 



Fig. 2 



ANALYTIC GEOMETRY 



PK 



2. In Fig. 2 the tangent of the angle POK is 



which 



we will represent by t. 

Now let the point P move continually along the line MN. 
Then PK and OK will change their values continually, and may 
be made to take an infinite number of different values in con- 
sequence of this change in the position of the point P. The 
tangent /, however, will always retain the same value during 
the operation of this change in the position of the point P, be- 
cause its angle POK does not change. 

Hence we call PK and OK variables, but we call / a con- 
stant. 




Fig-- 3 

3- In Fig. 3 let F' and F represent two fixed pins, to which 
the ends of an inelastic thread are fastened. At P letapencil 
be pressed against the thread. Let the point of the pencil be 
moved continually so as always to keep the thread stretched. 
The pencil will then trace out a curved line like APBDA. 

Let the length of the thread F'P + PF be represented by /, 
the angle PF'F by a, and the angle PFF' by b. 

Then F'P, FP and the angles a and b will change their 
values continually, and may be made to take an infinite num- 
ber of different values in conseouence of this chansre in the 
position of the point P, but /, the length of the broken line 
F'PF, and the length of F'F always retain the same value dur- 
ing the operation of this change in the position of the point P. 

We call F'P, FP, a and b va7dables, but we call / and F'F 
constants. 



LOCA TION OF POINTS IN A PLANE 3 

4. A Constant. — A constant quantity is one that always re- 
tains the same value throughout the operation of a given 
change. 

5. A Variable. — A variable quantity is one that may be 
made to take an infinite number of different values in conse- 
quence of the operation of a given change. 

6. The Change, — In analytic geometry the change which 
affects the values of the quantities investigated is generally 
the motion of a point along a given line or surface. 



CHAPTER II 

Location of Points in a Plane 

7. The position of a point in a plane may be indicated by 
means of its distances from any two fixed intersecting straight 
lines in the plane, these distances being measured parallel to 
the fixed lines. 



X' 





Y 




H 




P 

















K 




Y' 





X 



Fig. 4 

Thus, if YY' and XX' are any two fixed intersecting 
straight lines in the plane YOX, we may indicate the posi- 
tion of the point P in that plane by giving its distance HP 
from the line YY', measured parallel to XX', and its distance 
KP from the line XX', measured parallel to YY'. 



4. ANALYTIC GEOMETRY 

For convenience we will let the line XX' be horizontal. 

8. The Coordinate Axes. — The fixed intersecting lines 
YY' and XX' are called the axes of coordinates. 

9. The Axes of Abscissas and Ordinates. — The hori- 
zontal axis is called the X axis, or the axis of abscissas ; the 
other axis is called the Y axis, or the axis of ordinates. 

10. The Origin. — The point where the axes intersect each 
other is called the origin. 

11. Names of the Angles. — The four angles which the 
axes make with each other are called the first, second, third 
2Mdi fourth angles. 

They^r^-/ angle is the one above the X axis and to the 
right of the Y axis. 

The second angle is the one above the X axis and to the 
left of the Y axis. 

The third angle is the one below the X axis and to the 
left of the Y axis. 

Th^ fourth angle is the one below the X axis and to the 
right of the Y axis. 

12. The Coordinates. — The distances of any point from the 
axes, measured parallel to the axes, are called the coordinates 
of the point. 

13. The Abscissa. — That coordinate of a point which is 
parallel to the X axis is called the abscissa of the point. 

14. The Ordinate. — That coordinate of a point which is 
parallel to the Y axis is called the ordinate of the point. 

Thus in Fig. 4, PH is the abscissa of the point P. 
Since PH = OK, OK is often called the abscissa of P. 
PK is the ordinate of the point P. 

15. An abscissa is considered positive when it extends from 
the Y axis towards the right, and negative when it extends 
from the Y axis towards the left. 

16. An ordinate is considered positive when it extends from 
the X axis upwards, and negative when it extends from the 
X axis downwards. 



LOCATION OF POINTS IN A PLANE 



17. In giving the coordinates of a point we name the ab- 
.scissa of the point first. 

Thus when we say that the coordinates of a point are 
— 3> 5> we mean that its abscissa is — 3, and its ordinate is 5. 

Y 



P 



X'- 



Y' 

Fig. 5 

18. To locate this point in the plane YOX, we take an}^ 
convenient length, as a quarter of an inch, for our unit of 
length, and measure off, as in Fig. 5, 3 such units along the 
X axis from the origin towards the left, since the abscissa is 
— 3, and from the point thus reached measure off 5 units from 
the X axis parallel to the Y axis and upwards, since the ordi- 
nate is 5. Hence P is the point whose coordinates are — 3, 5. 

ig. Again, to locate the point w^hose coordinates are 4, — 6, 
we measure off 4 units along the X axis from the origin to the 
right, since the abscissa is positive, and from the point thus 
reached measure off 6 units parallel to the X axis and down- 
ward, since the ordinate is negative. Hence R is the point 
whose coordinates are 4, — 6. 

20. Similarly locate in the plane YOX the following points : 

The point whose coordinates are 4., 6. 
2 

3 
4 

5 

6 

7 



3, 


—5. 


b, 


2. 


7. 


4- 


0, 


5- 


4, 


0. 


0, 


0. 



6 ANAL YTIC GEO ME TR Y 

What are the signs of the coordinates of a point in the 
third angle? of one in the first angle? of one in the second 
angle ? of one in the fourth angle ? 



CHAPTER III 

The Construction of Loci 

21. By means of an equation we can often locate one or 
more series of points which together form a geometrical line 
or figure. 



KXAMPi^K I 



Take the equation x^ -\-y^ = 25. 
[i] x'+y' = 25, 



[2] 
[3] 



Then 
And 



y — 2s—x'' 



y = ±l/2S 



■X' 



Now in [3] let x^o, i, 2, 3, etc., successively, and find 
the corresponding values of y. Then let jr= — i, — 2, — 3, 
etc., and find the corresponding values of y. 



We will get 



[I] 
[2] 
[3] 
[4] 
[5] 
[6] 
[7] 



J^==t 5 

y — ^^ 
r = ± 3 

j= o 

y=^V~- 



■II 



when X 

X 
X 
X 
X 
X 
X 



o, 

I, 

2, 

3, 

4, 

5, 
6. 



It is 



We see from [7] that y is imaginary when ;r= 6, 
also imaginary for all values of x greater than 6. 

Now substituting the negative values of x in [3] we get 



THE CONSTRUCTION OF LOCI 



[8] 


J' = ± 4-8 




when X ^^ — i, 


[9] 


r = ±4-5 




" X— 2, 


[10] 


J>' = ±4 




" -^= 3> 


[11] 


^' = ± 3 




" ^ = —4, 


[.2] 


_j' = d= 




" -^ — — 5, 


[13] 


^ = ± l/- 


-II 


" jr = — 6. 



We see from [13] thatjF is imaginary when;i; = — 6, and 
that it is also imaginary for all values of x which are nega- 
tive and whose absolute values are greater than 6. 

Now taking the first set of values of x andjK, namely, x=ro 
andy^dzS, we can locate, by the method given in Sec- 
tions 18 and 19, two points on the axis of ordinates, one 5 
units above the axis of abscissas, and the other 5 units below 
it. Thus we get the two points A and B of Fig. 6. 



X' 







1 


1 


^^ 


'— 


Y 

A 


n 
1 


E 


G 








R 


/ 






• 


Co 






\ 


1 
\ 


\ 


T( 























\ 
K 


-5 


-4 

\ 


-3 
\ 


—2 


—I 


■ 


I 


2 


3 


4 


/ 


5 




K 


s 


1 B 


1 


) 


F 


















Y' 













Fiof. 6 



Taking the second set of values, namel}^ ;f = i andj' = 
±4.8, we can locate two more points C and D of Fig. 6. 

Taking the third set of values, namely, jt = 2 and,;^ =- ±4.5, 
we can locate two more points E and F of Fig. 6. 



8 ANALYTIC GEOMETRY 

Similarly, 

the 4th set of values gives us the points G and H, 
" 5th " " " " " '' '' I and J, 

" 6th '' " *' " " " point K. 



The 7th set of values, namely, ;ir=6 and j/ rz: ^ |/ — n, 
shows that it is impossible to locate, by means of this equa- 
tion, any point whose abscissa shall be — 6. 

Moreover, since jf is imaginary for all values of x greater 
than 6, it is impossible, by this equation, to locate any real 
point to the right of K. 

Again, 

by the 8th set of values we can locate the points L and M, 
" " 9th " " " '' '' " " " NandO, 
loth " " " " " '' " " P and Q, 
nth " " " " " " '* " Rand S, 
i2th " '* " " " " " point T. 



( t ( ( 



( ( ( ( 



( < t < 



The 13th set, namely, x =^6 andj|/=zhi/ — n, shows that 
it is impossible, by means of this equation, to locate any point 
whose abscissa is — 6. 

Since y is imaginary for all negative values of x whose 
absolute values are greater than 6, it is impossible, b}^ means 
of this equation, to locate any point to the left of T. 

By joining all the points located in this way we get the cir- 
cle AKBT. 

It is obvious, that if, in addition to the values of x used above, 
we take fractional values between them as o.i, 0.2, 0.3, etc. ; 
I.I, 1.2, 1.3, etc.; 2.1, 2.2, 2.3, etc.; points may be located 
as near to each other as we please, and the line AKBT can be 
made as nearly continuous as we please. 

EXAMPLK 2 

Take the equation xy = 10. 

[i] xy— 10, 

r -I ^° 

[2] J=-. 



THE CONSTRUCTION OF LOCI c 

If in [2] we let ;i; = o, i, 2, 3, etc., and find the correspond- 
ing values of y, we get 



[I] 


y r= 00 v\ 


"hen ;<; = 0, 


[2] 


jK = 10 


" r = I, 


[3] 


y— 5 


'' ^=2, 


[4] 


7= 3i 


" -^=3, 


[5] 


>= 4 


" -^ = 4, 


[6J 


_y = 2 


" -^ = 5, 


[7] 


J^= l| 


'• :r = 6, 


[8] 


J^= if 


" x=7, 


[9] 


y^ i\ 


" Jt; = 8, 


[to] 


y^ 4 


" ^ = 9, 


["] 


y— I 


" ;r := 10. 


If we let X r= 


I, 2, 3, 


etc., we get 


[12] 


jj/ = — 10 when X =^ — i, 


[13] 


JJ^= — 5 


" X— 2, 


[14] 


y = — z\ 


" x — — ^. 


[15] 


y— 2^ 


" ■^= 4, 


[16] 


jj/= 2 


" ■^= 5, 


[17] 


j^= if 


" X ^^ — 6, 


[18] 


j = - If 


" X— 7, 


[19] 


y—— li 


" ^ = — 8, 


[20] 


>^= 4 


" •^= 9i 


[21] 


j= — I 


" ;i; := — 10 



Now, from the second set of values, namely, xz=^\, and 
jj/=3 10, by the method given in Sections 18 and 19, we locate 
the point A in Fig. 7. 

By the third set, x^o. and jj^ = 5, we locate the point B. 

By the fourth set, .^t^ = 3 and_y = 3^, we locate the point C. 

By the fifth set we locate the point D. 

By the sixth set we locate the point E. 

Similarly, the remaining sets of positive values enable us 



lO 



ANALYTIC GEOMETRY 

Y 



X'' 




c\ 



Fig. 7 

to locate points to the right of the axis of ordinates and above 
the axis of abscissas. 

By joining these points we get the line ABCDE. 

In the same way the 12th, 13th, 14th, etc. sets of values 
enable us to locate points to the left of the axis of abscissas 
and below the axis of ordinates. 

Joining these points we get the line A'B'C. 

It is obvious, that if, in addition to the values of x used 
above, w^e take fractional values between them aso.i, 0.2, 
0.3, etc., and i.i, 1.2, 1.3, etc., we can locate points as near 
to each other as we please, and so make each of the lines 
ABCDE and A'B'C as nearly continuous as we please. 

The first set of values indicates that the point whose co- 
ordinates are ;t:=o, and jj/ = oo, is on the axis of ordi- 
nates at an infinite distance above the axis of abscissas. 
Hence the line EDCBA continually approaches the Y axis, 
can be made to approach as near to it as we please, but can 
never touch it. 

If in [2] we make ;r 1= co , we get jj/ = o. This set of 
values, indicates that the point whose coordinates are 
^=00 andjj/ = o is on the axis of abscissas at an infinite 
distance to the right of the axis of ordinates. Hence 
the line EDCBA continually approaches the X axis in the 
same way as it approaches the Y axis. 



THE CONSTRUCTION OF LOCI ii 

Show that A'B'C approaches the axes in the same way. 

KXAMPLK 3 
Take the equation i^y^ — 9;^':= — 144. 

[i] 
r2] 

[3] 
[43 



1 6/ 


1} 

— gx' = 


I44> 


i6y 


2 
= gx — 


144, 


f 


— 9 ^2 _ 

— 16-^ 


-9, 



y 



^l/xe-^' — 9. 



Now, in [4], if we let ;f = o, i, 2, 3, etc., and find the 
corresponding values of jj/, we get 



[i] 
[2] 
[3] 

[4] 
[5] 
[6] 
[7] 
[8] 
[9] 



J = ± v-_ 
^ = ±1/- 
^ = ±1/: 

y ^ o 
y = ± 2i 

JJ/=:dz3.3 

j/ = zb4-3 
J = d= 6 



; 7 

'l"S" 



31 5 



when X ^o, 
jf = I, 

X ^ 2, 
X— Z, 
^ = 4, 

X— 5, 
;c = 6, 

Ji:= 8. 



If in [4] we let x = — i, — 2, — 3, etc., and find the cor- 
responding values of jK, we get 



[10] 

[11] 
[12] 

[13] 
[14] 
[15] 
[16] 

[17] 



y = ±i/-8tV 


when 


X — I, 


y = ±:i/-6| 




^ — 2, 


y = ±V—3ii 




-^— 3, 


J/ = 




X — 4, 


r = d= 2i 




■^— 5, 


>' = :t 3.3 




X = — 6, 


y = ±: 4.3 




^_ 7, 


y^^ 5.2 




;r = —8. 



12 



ANALYTIC GEOMETRY 

Y 




Fig. 8 

The first four sets of values show that for this equation y 
is imaginary when jr ^ o, i, 2, or 3, and hence it is impossi- 
ble, by means of this equation, to locate any real point whose 
abscissa shall be o, i, 2, or 3. 

The fifth set of values, namely x^ 4. and j^ = o, enables us 
to indicate the point A on the axis of abscissas to the right of 
the axis of ordinates. 

The 6th set of values locates the points B and C, 
" 7th " " '* ** '' " Band K, 

" 8th '' " " '* " " Fand G, 

" 9th " " " " ** " H and I. 

It is obvious that every set- of values of x greater than 8 
will give two values of jk, which will be equal, but which will 
have opposite signs, and hence that we may locate points 
farther and farther to the right. 

Joining the points located we get the line HDAEG. 

The loth, nth, and 12th sets of value show that for this 
equation y is imaginary when jr = — i, — 2, or — 3, and hence 
that it is impossible, by means of this equation, to locate any 
real points whose abscissas are — i, — 2, or — 3. 

The 13th set locates the point K on the axis of abscissas to 
the left of the axis of ordinates. 



THE CONSTRUCTION OF LOCI 13 

The 14th set locates the points L and M. 

Similarly each of the other sets enables us to locate two 
points to the left of the axis of ordinates, and at equal dis- 
tances above and below the axis of abscissas. 

It is obvious that every negative value of x' whose absolute 
value is greater than 8, will give us two values of y, which 
will be equal, but which will have opposite signs, and hence 
that we may locate points farther and farther to the left. 

Joining the points located we get the line PI^KOQ. 

It is obvious, that if, in addition to the values of y used 
above, we use the fractional values between them, we can 
locate points as near .to each other as we please, and so make 
each of the lines HDAEG and PLKOQ as nearly continuous 
as we please. 

Draw the lines represented by the following equations : 

[4] 2>y — ^x— o, 

[5] 3/ + 4-^ = o» 

. [6] 4/ = 3-^'> 

[7] loy = x^ — X — 20, 

[8] y - x\ 

. [9] y — X —\x\ 

[10] y'^ = \ox, 

[11] i6y^ + 9Jt:' — 1441=0, 

[12] i6jj/' — 9Jt:' -|- 144 = o. 

In the next three examples substitute the values of x before 
expanding the right hand members. 

[13] y = C-^ — 2) (-r — 7)-, 

[14] y = (-^— 3) (-^— 7) (-^— 11), 

[15] /= (-^— 3) (-^—7) (■^— II) (■^— 13)- 

22. It is obvious from the method by which the points were 
located in the line obtained by means of the equation of Ex- 
ample I, that all points on the line AGKHBQTP, in Fig. 
6, are fixed in position by one and the same law, namely, 
that the square of the abscissa plus the square of the ordinate 



14 ANAL YTIC GEO ME TR Y 

is always equal to 25. For if in each of the sets of values of x 
and J/ taken to locate the points, we square the value of ;rand 
also of jK, and add them, we will always get 25. 

23. The line AGKHBQTP is called the locus of the equa- 
tion x^-|-jj/^= 25. 

24. It is also obvious from the method by which the points 
were located in the lines ABC and A'B'C of Fig. 7, obtained 
by means of the equation of Example 2, that all the points 
on those lines are fixed in position by one and the same law, 
namely, that the abscissa, multiplied by the ordinate, is 
always equal to 10. For if in each of the sets of values of x 
andjj/, taken to locate those points, we multiply the value of 
X by that of jk, we get 10. 

25. The lines ABC and A'B'C together are called the locus 
of the equation ;rj/ = 10. 

26. Similarly it may be shown that all the points on the 
lines HAI and PKQ of Fig. 8, obtained by means of the 
equation of Example 3, are fixed in position by one and the 
same law, namely, that 16 times the square of the ordinate 
minus 9 times the square of the abscissa is always equal to 
— 144. 

27. The lines HAI and PKQ together are called the loctis 
of the equation i6y'^ — gx' = — 144. 

EXERCISK 

28. Give the law according to which the points are fixed 
in position in each of the lines obtained by means of each of 
the equations given above, namely, equations 4, 5, 6, 7, etc. 

Hence we may define a locus as follows : 

29. A Locus. — A locus is the whole assemblage of points , each 
of which is fixed in position by one and the same law. 

30. It is obvious, that if in Example i, page 6, in addition 
to the values of x and jj/ there used, we consider the fractional 
values between them, then there are an infinite number of sets 
of values of x 2LVidLy, satisfying the equation corresponding to 
an infinite number of points on the locus. Hence if the x and 



777^ CONSTRUCTION OF LOCI 15 

y of the equation represent a point P on the locus KGANR, 
etc., then by moving the point P along the locus, we may 
make the x andjj/ of the equation take an infinite number of 
different values. 

31. Hence the x and y of the equation x:' -{-y^ = 25 are 
variables. by § 5. 

32. The equation x' -\-y^ = 25 is called ^/le equation of the 
locus KGPSJK (Fig. 6). 

33. It is also obvious, that if in Example 2, page 8, in addition 
to the values of x and jk there taken, we consider the fractional 
values between them, then there are an infinite number of sets 
of values of x and y satisfying the equation corresponding to 
an infinite number of points on the locus. Hence if the x and y 
of the equation represent the coordinates of a point P on the 
locus, then by moving the point P along the locus we may 
make the x andy of the equation take an infinite number of 
different values. 

34. Hence the x and y of the equation jrj/ = 10 are varia- 
bles. . b}^ § 5. 

35. The equation xy = 10 is called the equation of the 
locus ABCA'B'C (Fig. 7). 

36. In the same way we may show that by moving the 
point P along the locus of the equation of Example 3, the x 
andjK oi that equation may be made to take an infinite num- 
ber of different values. Therefore the x and y of the equa- 
tion i6jK^ — <^x^ = — 144 are variables. by § 5. 

37. The equation \6y^ — 9;t:^= — 144 is called the equation 
of the locus HBAIPKQ (Fig. 8). 

38. It is obvious, from the method by which each of the 
loci in Examples 1,2, and 3 was drawn, that the values of 
the X and y in each of the sets of values which satisfy the 
equations of the loci, are so related to each other that 

ist, the X and y of each set of values of the variables 
which satisfy the equation stand for the coordinates of a par- 
ticular point on the locus, and 



i6 ANALYTIC GEOMETRY 

2d, the values of the coordinates of each point on the locus 
satisfy the equation of the locus. 

Hence we may define the equation of a locus as follows : 

39. The Equation of a Locus. — The equation of a locus is 
one in which the variables stand for the coordinates of every 
point on the locus. 

40. Corollary I. — The values of the coordinates of every point 
on a locus must satisfy the equation of a locus. 

41. Corollary 2. — If the values of the coordinates of a point sat- 
isfy the equation of a locus, that point miist be 07i the locus. 

by § 38, ist 

Measurement of Arcs 

42. An arc of any circle whose length is equal to the length 
of the radius of that circle is often taken as the unit for meas- 
uring arcs of that circle. 

[i] The circumference = ttD = 2 7rR by Geom. 29. 

Now, when the length of the radius is taken as the unit for 
measuring the lengths of arcs, [i] becomes 

[2] The circumference =: 2 7r, 

and 

[3] The semi-circumference =: 7r= 3.1416. 

That is, the length of the semi-circumference of any circle 
is equivalent to 3.1416 units when each unit is as long as the 
radius of the circle. 

Hence, when the length of the radius is taken as the unit 
for measuring arcs 

[4] 180° = 3.1416, 
[5] o° = o, 

[6] 10' = 0.17, 

[7] 20° = 0.35, 

[8] 30° = 0.52, 



MEASUREMENT OF ARCS 17 



[9] 40° = 0.70, 

[10] 50° = 0.87, 

[11] 60° = 1.05, 

[12] 70° ^= 1.22, 

[13] 80° = 1.40, 

[14] 90° = 1.57, 

[15] 180' = 3.14, 

[16] 190'' = 3.31, 

[17] 200° = 3.49, 

L18] 210° = 3.66, 

[19] 220° = 3.84, 

[20] 230° = 4.01, 

[21] 240° = 4.19, 

[22] ^ 250° — 4.36, 

[23] 260° = 4.54, 

[24] 270° = 4.71. 

Corollary. — Since by equation 2 



[l] 


2 7t z^ 360'', 


[2] Then 


7t — 180°, 


!3] and 





EXAMPIvES 

Draw the curves of sines whose equation is 
[i] jK= sin X. 

For the values of the arc x in this equation take the series 
of values given in the second column of the table in § 42. 

For the values of j/ take from the Trigonometrical Tables the 
natural sines of the number of degrees corresponding to each 
of these values of x. In this way we will get the following 
sets of values of x and y : 



i8 



ANALYTIC GEOMETRY 



When jr = o, 

X =. 0.17, 
^ = 0.35, 

X =: 0.52, 

X = 0.70, 
X = 0.87, 
X = 1.05, 

X = 1.22, 

X = 1.40, 

•^ = i.57» 

^ = 3-14, 

^ = 3-31, 
X = 3.49, 

X = 3.66, 

^ = 3.84, 
X = 4.01, 
:r = 4.19, 
X — 4.36, 
X = 4.54, 

.^ = 4.71, 



y — o, 
y — 0.17, 
y = 0-34, 

JV = 0.50, 
J/ = 0.64, 
J/ — 0.77, 
y = 0.87, 
jj/ = 0.94, 
y = 0.98, 

j; z= 1. 00, 

_y = 0.00, 

J/ = — 0.17, 

;j/ = —0.34, 

y — — 0.50, 

JK = — 0.64, 
y — — 0-77, 

J^ = — 0.87, 
y = —0.94, 
jv == — 0.98, 
_y = — 1. 00, 



If with these values of x and j/ we locate a series of points 

by the method given in §§ 18 and 19, we will get the following 

locus : 

Y 




Fig. 9 

2, Draw the curve of tangents whose equation is 

y = tan x. 



MEASUREMENT OF ARCS 



19 



For the values of the arc x in this equation take the series 
of values given in the second column of the table in § 42. 

For the values of j' take from the Trigonometrical Tables the 
natural tangents of the number of degrees corresponding to 
each of these values of x. In this way we get the following 
sets of values for x and y : 



When ;t: := o. 

;t: = ± 0.17, 
X — :^ 0.35, 
^ = zb 0.52, 
;>; = rh 0.70, 
*;f = d= 0.79, 
Jt: = zb 0.87, 
X — ±1 1.05, 
;»; = zb 1.22, 
;i; = rb 1.40, 
X — ±: 1.57, 



y — o, 

y — ±, 0.18, 

y — ^ 0.36, 

^^ = ±0.58, 

J = db 0.84, 

>' = zb 1. 00, 
J = rb 1. 19, 
JJ/=zb 1.73, 

y — :^ 2.75, 
y — ^ 5.67, 



*x = zb 0.79 is the value of 45° obtained as in § 42. 

If with these values of x andji^ we locate a series of points 
by the method given in §§ 18 and 19, we will get the following 
locus. 

Y 




3. Draw the curve of secants whose equation is 

y z= sec X. 



CHAPTER IV 



The Intersection of Loci 



KXAMPI^KS 



43. Where does the locus of 167^ + gx'^ — 144 = o cut the 
Y axis ? Where does it cut the X axis? 




From Example 1 1 , p. 13, we learn that A'BAB' is the locus of 

[i] i6y' + 9^' — 144 = 0. 

Since the point B is on the locus, its coordinates must sat- 
isfy the equation of the locus. by § 40. 
But the coordinates of B are x = o andjK= OB. 
Substituting these values for the x andjj/ of [i] we get 



[2] 


16. OB 


144 = 0, 


[3] or 




i6.0B= 144, 


[4] 




OB = 9, 


[5] 




OB = ±3. 



THE INTERSECTION OF LOCI 21 

Hence i6r^ -j- qjt^ — 144 = o cuts the jf axis in two points, 
one above and the other below the origin, and both at a dis- 
tance of 3 units from it. 

Since the point A is on the locus, its coordinates must sat- 
isfy the equation of the locus. by § 40. 

But the coordinates of A are x = OA and jk = o. 
, Substituting these values of x and j/ into [i] we get 

[6] 9.0A'= 144, 

[7] oa'= 16, 

[8] OA = d= 4. 

Hence 167^ + 9^^ — 144 = o cuts the X axis in two 
points, one to the right and the other to the left of the origin, 
and both at a distance of 4 units from it. 

44. Where does the locus of i6j/' — 9-^' + 144 ^ o cut the 
X axis ? Where does it cut the Y axis ? 

Let A, Fig. 12, represent the point in which the locus of 
this equation cuts the X axis. 

Then since A is on the locus, its coordinates must satisfy 
the equation of the locus. by § 40. 

But the coordinates of A are x = OA and jk = o. 

Substituting these for the x and y of equation 

[i] 16/— 9x'-f 144 = o, 

we get 

[2] — 9.0A = — 144, 

[3] OA = 16, 

[4] OA = ± 4. 

Hence this locus cuts the X axis in two points on opposite 
sides of the origin, and both at a distance of 4 units from it. 

Let B, Fig. 12, represent the point where the locus is sup- 
posed to cut the Y axis. 

Then if B is on the locus, its coordinates must satisfy the 
equation of the locus. by § 40. 



22 



ANAL YTIC GEO ME TR Y 



But the coordinates of B are jf = o and j^ = OB. 
Substituting these values for the x and^ of [i] we get 



[5] 

[6] 
[7] 



16. OB = — 144, 

OB = — 9, 

OB = zb 1/ — 9. 



Equation [7] shows that it is impossible for this locus to 
cut the Y axis. 

From Example 3, p. 11, we learn that the locus of [i] is 
LAKMA'N in Fig. 12. 





r 

Fig. 12 , 

Hence we may devise the following 

45. Rule. — To find where any locus cuts the X axis, make 
jF = o in the equation of the locus, and the values of x obtained 
from the equation will be the absciSvSas of the cutting points. 

For, the values of the coordinates of the cutting point must 
be J/ = o, and x = the abscissa of the cutting point. 

But since the cutting point is on the locus, the values of 
these coordinates must satisfy the equation of the locus. 

by § 40. 

Hence, if we substitute these into the equation of the locus 
the terms involving y will disappear and leave an equation 



THE INTERSECTION OF LOCI 



23 



whose only unknown quantity is x, the abscissa of the cutting 
point. 

Similarly, we may devise the following 

46. Rule. — To find the point where any locus cuts the Y axis, 
make ;t: := o in the equation of the locus, and the values of y 
obtained from the equation will be the ordinates of the cutting 
points. 

EXAMPLEvS 

Where do the loci of the following equations cut the axes ? 



1. Jt:-+jF =9. 

2. 2^y' -\- l6x^ := 400. 

3. 25j^ — i6;»;' = 400. 

4. y^ ^ \ox. 

5. y = {x — 2) (^—7) 
6 



Ans. jr := zt 3, JF = o. 
y — ^^,x — o. 

Ans. jc = zb 5, J^' = o. 

jj^ = it 4. -^ = o- 

Ans. ^ = zb 5, y = o. 

Ans. X ■= O^ y z:zz o. 

Ans. j<; = 2 and 7. 
y=(jf— 3)(;>;— 7)(;t:— ii)(;r— i3).Ans.:r=3,7, 11,13. 
y. y"^ = (jt: — a) (jr — <^) (;r — <;) . Ans. x ^ a, b and r. 



EXAMPI.ES 

47. Where does x'^ -\- y~ = 25 cut xy^ 10 ? 




From Example i, p, 6, we see that ABD is the locus of 

[i] ^'+/=25. 



24 ANAL YTIC GEO ME TR Y 

From Example 2, p. 8, we see that RPSTU is the locus of 

[2] xy = 10. 

lyet x' and y' be the coordinates of the point P. 
Then, since the cutting point P is on the locus ABD, its 
coordinates must satisfy the equation of this locus. by § 40. 
Hence, substituting them into [i] we get 

[3] ^"+y^ = 25. 

Since P is also on the locus RPSTU, its coordinates must 
satisfy the equation of that locus. by § 40. 

Hence, substituting them into [2] we get 
[4] xy = 10. 

Now, since ^'= OK andj/' = PK in both [3], and [4] these 
equations are simultaneous and may be solved by algebra. 
From [4] we get 



[5] 




Substituting 


this value of y' into [3] we get 


[60 


, 100 

^ + ^n = 25, 


[7] 


X^^ 2^X^^ = 100, 


[8J 


,4 ,., - 625 22s 

x'' — 25X" + ^ — ^ , 
4 4 


[9] 


^'^ = 20 or 5, 


[10] 


^' = rb 4.5 or d 



Substituting these values of x^ separately into [3] we get 
[11] r' =^ 2.2 when ;t:' = 4.5, 



[12] 


y ■=i — 2.2 


'* x' — - 


-4.5. 


[13] 


y = 4.5 


" x' — 


2.2, 


[14] 


y = 4.5 


" x' — - 


— 2.2. 



Equations [11], [12], [13], and [14] show that the curves 
cut in four points, P, S, F, and G. 



THE INTERSECTION OF LOCI 25 

Hence we may derive the following 

48. Rule. — To find where two curves cut each other, treat 
their equations as simultaneous and solve them. The values 
X and y thus found will be the coordinates of the cutting 
points. 

For, since as in Fig. 13 the cuttingpointPison both curves, 
its coordinates must satisfy the equation of both curves. 

by § 40. 

Hence we may let the abscissa of the cutting point be rep- 
resented by x\ and the ordinate of the cutting point be rep- 
resented by y in the equation of both curves. Then the equa- 
tions will be simultaneous and may be solved by algebra. 

The values of x^ and y\ obtained by solving these equa- 
tions, will be the coordinates of the cutting points. 

EXAMPLES 

Solve each of the following examples and illustrate it by 
drawing its locus. 

1. Where does 7 = ;r -j- 5 cut _y = ;r'^ ? 

Ans. :r = 2.8, jj/ =^ 7.8. 

■^ = 3-2,jK = — 1.8. 

2. Where doesjj/ = 2x q.\x^. xy ^= 18? Ans. .r = 3, _y = 6. 

x = — 3,jF= — 6. 

3. Where does x'^ -\-y'^ = 25 cut jj/ ^= x- ? 

Ans. ;»; = dz 1.5, JK ^ 2.2. 

4. Where does jk' = 4X cut ;i;^ -f-jj/^ = 25 ? 

Ans. ^ = 3.4, jj/==dz3.7. 

5. Where does j = x^ cut ^y^ = s^'^ ? 

Ans. x = l,y=dzj\. 
x^ o,y=o. 

6. Where does x^ -\-y' ^ 16 cut 36)/^ + gx^ r= 324 ? 

Ans. 

7. Where does looy^ + 6/^x^ ^= 6400 cut 6^y^ — 36;!;^ = 
— 2304 ? Ans. 



CHAPTER V 

The Straight Line 

49. The Intercept. — The intercept of any Hne on either 
axis is the distance from the origin to that line measured 
along that axis. 

50. Every line which cuts the X axis makes four angles 
with it. Starting on the axis in the positive direction from 
the common vertex of these angles and moving counter clock- 
wise, the angles are named first, second, third, and fourth. 

Y 



r- 




r 



Fig 14 



51. The Inclination. — The inclination of a line is the first 
of the four angles which it makes with the X axis. 

52. The Slope. — The slope of a line is the tangent of its in- 
clination. 

PROPOSITION I 

53. The equation of a straight line in terms of its slope and Y 
intercept is 

y = sx-\-b, 

in which s is its slope and b its Y intercept. 

Let YY' be the axis of ordiiiates and XX' the axis of ab- 
scissas, and let MN be any straight line in their plane. 

Let ^ = tan BRH, the slope of MN ; and ^ = 0B, its Y in- 
tercept. 



THE STRAIGHT LINE 

Y P/N 



27 




Fig. 15 

Let P be any point on the line and draw its coordinates PH 
and OH. 

Let ;t:=OH andjF = PH. 
We are to prove that 

y =:i SX -\^ b 







is the 


equation of MN. 


Draw BK |1 OH 


. 




[i] 




z:pbk = z:1prh, 


by Geom. 8. 


[2] 


hence 


tan PBK = tan PRH = 


>?, 


[3] 




PTC 
tan PBK =—^, 


by Trig. 3. 


[4] 


hence 


PK 
BK ■"' 




[5] 


and 


PK-=^.BK. 




L6] 


But 


BK = OH, 


by Geom. 17. 


[7] 


hence 


PK = 5.0H. 




[8] 




KH-=BO. 


by Geom. 17. 


By adding [7] 


and [8] we get 




[9] 




PK + KH = 5.OH + BO, 




[10] 


or 


y^sx-\-b. 





Now the X andj^ of [10] stand for the coordinates of the 
point P. But, since P is any point on the line MN, we 



28 ANALYTIC GEOMETRY 

ma}' move it either way as far as we please along that line. 

Hence the x andjK of [lo] stand for the coordinates of every 
point on the line MN. 

By moving the point P along the line, we may make the x 
and y of [lo] take an i7ifinite number of different values. 
They are therefore variables. by §5. 

But s and b always retain the same values as the point P 
moves along the line MN. They are therefore constants. 

by §4. 
Therefore, since the variables in [10] stand for the coordi- 
nates of every point on the line MN, [10] must be the equa- 
tion of the line MN. by § 39. 

Q. K.D. 

54. Corollary i. — Any equation of the form 

y zzr co7istant.x-\- constant, 

is the equation of some straight line. 

For, since MN may be any straight line, its intercept b 
may have any value from o to + co , or to — 00 . 

Hence b may be any constant whatever. 

Also, since MN may be any straight line, its inclination 
may have any value from 0° to 180°, and therefore its slope 
may have any value from o to + 00 , from o to — 00 . 

Hence s may be any constant whatever. 

Therefore, y ^^ sx -\- b includes every equation of the form 
y = const. :r + const. 

Hence every equation of that form must be the equation of 
some straight line. 

55. Corollary 2. — The equation of a straight line passing 
through the origin is 

y = sx. 

For, for every such line /^ = o, and [10] becomes 

_y =r SX. 

56. Corollary ^. — The equation of a straight line parallel to 
the X axis is 

y ^= b. 



THE STRAIGHT LINE 



29 



For, for every such line the inclination must be 0° or 180°, 
and hence the slope must be o, by Trig. 19, and [10] must 
become 

Similarly, it may be shown, that if c be any constant, then 

:r = ^, 
is the equation of a line parallel to the Y axis. 

EXAMPLES 

1. What is the inclination and slope of the line y := 
\x-\- ^t What is its Y intercept? 

Its slope = \^ 

or the tangent of its inclination = | , 

hence its inclination = 26° 34'. 

2, What is the inclination of each of the following lines 



by § 53- 
by § 52, 



y — To-^ + 2. Ans. 30° 58'. 

y— ^ + 3- Ans. 45". 

y=i 3^+1. Ans. 71° 34'. 

3. Which of the four angles made by the axes does the line 
J/ =^ — 2J»; + 5 cross ? 



X' 







Y 






1VI\ 




B 








2 


oX 


^^^-<f^» 






3 


4 

V 


"^ 


^N 



Fig. 16 



The Y intercept of this line is 5. by § 53, 

Hence the line cuts the Y axis 5 units above the origin, 
at B. 

The slope of the line is — 2. by § 53, 

Hence the inclination is an obtuse angle. by Trig. 16, 



30 



ANALYTIC GEOMETRY 



Therefore, y = — 2x-\- ^ must lie across the first of the 
four angles made by the axes, as MN in Fig. i6. 

4. Show which of the four angles each of the following lines 
crosses : 

J—-—2X — S, 

jj/=2X — 5, 

y= 2.r+5. 

5. Show how each of the following lines lies : 

J/ = 2X, 

r = — 5, 
)/ = 6. 

6. What are the intercepts of y =: ^x-\- 10 on the two 
axes ? 

7. What is the area of the triangle between y^=z^x-\- 10 
and the two axes ? 

8. What is the area of the triangle between _y = — ^x — 10 
and the two axes ? 

9. What is the area of the riangle between y := — iox-\- ^ 
and the two axes ? 

PROPOSITION II 

57. The equation of a straight line passing through a fixed 
point is 

y — y =: s {x' — x) , 

in which x' afidy' are the coordinates of the fixed point, x and y 
the coordinates of any poi?it on the line and s is the slope of 
the line. 




Fig. 17 



THE STRAIGHT LINE 31 

lyCt YY' be the axis of ordinates and XX' the axis of ab- 
scissas. 

Let P' be the fixed point and MN any straight line passing 
through that point. 

Let P be any point on the line MN and draw the coordi- 
nates of P and P'. 

Let ^ = OH and y = PH , 

x'=OH' '* y=P'H'. 

Let s = tan PRH, the slope of MN. 

We are to prove that 

y' — y = s {x' — x) 

is the equation of MN. 
Draw PK |{ XX'. 

^P'PK = ^PRH, byGeom. 8. 

hence tan P'PK — tan PRH — ^. 

tan P'PK =:r 1^, by Trig. 3. 



[I] 




[2] 




[33 


hence 


[4] 


But 




hence 


[5] 


and 


[6] 




[7] 





P'K_ 
PK""""' 
P'K = .y.PK:. 
P'K = P'H'— PH =y--y, by Geom. 17. 
PK = OH' — OH — x' — x, by Geom. 17. 

Substituting these values of P'K and PK into [5] we get 

[8] y'—y = s{x'~x). 

Now the X and y of [8] stand for the coordinates of the 
point P. But since P is any point on the line MN, we may 
move it either way as far as we please along that line. 

Hence the x andjF of [8] stand for the coordinates of every 
point on the line MN. 

By moving the point P along the line MN, we may make 
the X andy of [8] take an infinite number of different values. 
They are therefore variables. by § 5. 

But s, x\ and J'' always retain the same value as the point P 



32 ANALYTIC GEOMETRY 

moves along the line MN. They are therefore constants. 

by §4. 
Therefore, since the variables in [8] stand for the coordi- 
nates of every point on the line MN, [8] must be the equation 
of that line. by § 39. 

Q. E. D. 
EXAMPI^ES 

1. What is the equation of the line whose inclination is 45° 
and which passes through the point whose coordinates are 
—5, 10? 

Since this is a straight line passing through a given point, 
its equation must be of the form 

[i] y—y—s{^x' — x), by §57. 

in which ^ = tan 45"^ =1, by Trig. 26. 

;r' = — 5 and y == 10. 

Substituting these values into [i] we get 

[2] ' y—10— I (^ + 5). 

Simplifying this equation we get 

[3] y—x-\-\^, 

which is the equation of the line passing through the 
point — 5, 10, and whose inclination = 45°. 

2. What is the equation of the line whose inclination is 
45°, and which passes through the point whose coordinates 
are — 5, — 10 ? 

What intercepts does it cut from the axes ? 

Ans. y ^=^ X — 5. 

Y intercept = — 5, 
X intercept = 5. 

3. What is the area of the triangle between the axes and 
the line whose inclination is 135°, and which passes through 
the point whose coordinates are — 2, — 5 ? 

Ans. JK = — X — 7. 
Area of the triangle = 24J. 

4. Where does the line whose slope is 10, and which passes 
through the point whose coordinates are i, 5, cut the line 



THE STRAIGHT LINE 



33 



whose slope is — 2, and which passes through the point whose 
coordinates are — i , — 5 ? 

PROPOSITION III 

58. The equation of a straight line passing through two fixed 
points is 

in which x' and y^ are the coordinates of one of the fixed poi7its, 
x" and y are the coordinates of the other ^ and x and y the coor- 
dinates of a7iy pohit on the line. 




Let YY' be the axis of ordinates and XX' the axis of ab- 
scissas. 

Let P' and P" be the two fixed points, and MN a straight 
line passing through them. 

Let P be any other point on the line MN, and draw the co- 
ordinates of P, P' and P". 



Let-r =OH andjK = PH, 
x' =OH' " y =P'H', 
y = OH" " y'=P"H". 



We are to prove that 



y-^=€^!(^'-^). 



X — X 



is the equation of MN, 



34 ANALYTIC GEOMETRY 







Draw PK and P'K' || XX'. 




[l] 


Now 


^P"P'K' =:z:p'pk. 


by Geom. 8. 


[2] 


Hence 


tan P"P'K' = tan P'PK. 




[3] 




tanP"P'K'=^, 


by Trig. 3. 


[4] 




P'K 
tanP'PK = -p^, 


by Trig. 3. 


[5] 


hence 


P"K'_P'K 
P'K' — PK' 


by [2]. 


[6] 


and 


P"K' 
P K — p,^, PK. 




[7] 


But 


P'K=y->', 




[8] 




p"K'=y' y, 




[9] 




P'K' = ^"— < 




[10] 




PK r=;t'' — ;r, 





/' 

^—X' 



y" — y' 
[11] Hence y'—y^-'-j, — ^,{x' — x). by [6]. 



Now the X and j of [11] stand for the coordinates of the 
point p. Since P is ' any point on the line MN, we may 
move it either way as far as we please along that line. 
Hence the x andjF of [11] stand for the coordinates of every 
point on the line MN. 

By moving the point P along the line MN, we may make the x 
and jK of [11] take an infinite number of different values. 
They are therefore variables. by § 5. 

But x', x'',y andj/" always retain the same values as the 
point P is moved along the line MN. They are therefore con- 
stants, by § 4. 

Therefore, since the variables in [11] stand for the coordi- 
nates of every point on the line MN, [11] must be the equation 
of MN. by § 39. 

Q.K.D. 

y" — y' 
59. Corollary i. — The fraction— ^^ — —^ is the slope of the line 

which passes thro2igh the two fixed points whose coordinates a7'e 
x\ y\ and x", y". 



THE STRAIGHT LINE 35 

For 
[i] tan P"P'K' = 1^, by [3], §58. 

[2] but tan P"P'K' = tan PRH, by Geom. 8. 

P"K' 
[3] hence tan PRH = -p^- 

[4] But — -^ -^ 



P'K' x'' — x 



" ^M 



[5] hence "^^7^ — =^ := tan PRH = the slope, by § 52. 

60. Corollary 2. — The length of the line joining any two given 
points is 

in which x", y" and x\ y' are the coordinates of the two given 
points. 

For, in Fig. 18 



[i] P'P" = P"K' + P'K', by Geom. 26. 



[2] hence P'P" =(/'—/)'+ (■^" — ■^')', 



[3] or P'P" = -/(y'-y)^+(^"_;r')\ 

EXAMPI.ES 

I. What is the equation of the line which passes through 
the two points 3,5, and — 2, — 7 ? 

The equation of this line must be of the form 

[i] y_^:=^_^^(y_^), by §58. 

in which 

x' — z andy = 5, 
;t:" =r — 2 and jj^" = — 7. 

Substituting these values into [i] we get 
[2] ^_5 = Z1I^(^_3). 



Simplifying this equation we get 



36 



[3] 



ANALYTIC GEOMETRY 






which is the equation of the line passing through the two 
points 3, 5 and — 2, — 7. 

2. What is the equation of the line which passes through 
the two points — 5, 8 and 3, — 10 ? 

Ans. jj/= — 2\x — 3^. 

3. Where does the line which passes through the points 
— 3, — 5, and 5, 8 cut the axes ? 

Ans. At the point y^g, — 1, 

4. Where does the line which passes through — 5, 3 and 
3, — 5 cut the line which passes through 5, 5 and the origin? 

Ans. At the point — i, — i. 



PROPOSITION IV 



61. T/ie tangent of the angle betwee?i two straight lines is 
given by the equation 

c' c 

tan cp = 



] + s's' 

in which cp is the angle between the two Ihies, s' is the slope of one 
of the lines and s is the slope of the other. 




Fig-, 19 



Let MN and RS be any two straight lines. 

[i] Let^ = ^HPK. 

Let j'= sx-\- b be the equation of the line MN. 

[2] Then 5 ==: tan PHX. 



by §53" 



THE STRAIGHT LINE 37 

I^et JK = s^x -\- b' be the equation of the line RS. 

[3] Then ^ = tan PKX. by § 53. 

We are to prove that 

s' ^ 

tan (p~ , . 

I -|- s s 

[4] ^PKX=zlPHK4-^HPK, byGeom. 13. 

[5] hence ^HPK = ^ PKX — ZPHK, 

[6] and tan HPK = tan (PKX — PHK), 

r -1 1 / * ^v tan A — tan B , ^ . 

[7] but tan (A — B) = — j -, by Trig. 17. 

^'-' ^ ^ I + tan A tan B ^ ^ ' 

Since the A and B of [7] may be any angles whatever, let 
A = PKX and B = PHK. 

Substituting these values of A and B into [7] we get 

r..n /-r^T^^^ -r^-r-r-r^X taUPKX taUPHK 

[8] tan (PKX — PHK) ~ — . — — t^tttf- 

*- -• ^ ^ I + tan PKX tan PHK 

Comparing [6] and [8] we get 

tan PKX— tan PHK 



[9] tan HPK — 



I + tan PKX tan PHK" 



Substituting the values ol tan HPK, tan PKX, and 
tan PHK found in [i], [2] and [3] into [9] we get 



[10] tan cp = 



s' — s 



I + s's' 

Q.E.D. 



62. Corollary i. — Whenever two straight lines are perpendic- 
ir to each other 

i-\-s^s^=^o, 

in which s' is the slope of one of the two lines aiid s is the slope of 
the other. 

For when RSJ_MN 

[i] ^HPK = 90°, 

[2] and tan cp = tan 90° = oo , by Trig. 19. 

s' — s 
[3] hence , ,^ = co , by [10.] 

I -]- -f ,> 



38 ANALYTIC GEOMETRY 



s' — ^ 



But when the value of the fraction — _. , is infinitely large 
its denominator is infinitely small. 
[4] Hence i -[- ^'^ ;= o. 

63. Whenever two straight lines are parallel to each other 

s' — s=: o, 

i?i which j-' is the slope of one of the two lines and s is the slope of 
the other. 

For when RS || MN 
[i] 5'=^, by Geom. 8. 

[2] Hence 5' — s^o. 

EXAMPLES 

1. What is the angle between the two lines jk= 3^ + 1 and 

Let ^'=3, the slope of the first line, and ^ = |^, the slope of 
the second line. 

Substituting the values of ^ and s^ into [10], § 61 we get 

Hence the angle between the lines = 45°. 

2. What is the angle between the lines _y=3;r+ i, and 
y^=x — 5? Ans. 26° 34'. 

3. What is the angle between the lines y =^ x — 5, and 
y^ — ;t: + 2 ? Ans. 90°. 

4 . What is the angle between jj^ = 3-^ + 2 , and y=^\x} 

Ans. 45° 

5. What is the angle between jf = j-V-^> andjF = fjr? 

Ans. 32° 56' 36". 



THE STRAIGHT LINE 
PROPOSITION V 



39 



64. The equatioji of a straight line in terms of its intercepts on 
both axes is 

X . V 

in which a is its X intercept and b is its Y intercept. 









Y 






W-s 


^ 


R 


p 


Y'— 









^^<f^ 


A 








^N 








r 





Fig. 20 

Let MN be any straight line and P any point on it. 

Let ^ = OT and b = OR, 
:r = OK " j/ = PK. 

We are to prove that the equation of MN is 









X y 
a^ b^^' 




[I] 




KT 


: PK : : OT : OR, 


by Geom. 25 and 31 


[2] 


or 




a — X \y : : a : b. 




[3] 


Hence 




ab — bx =^ ay, 




[4] 


or 




bx -\- ay = ab. 




[5] 


hence 




X . V 





As in Propositions I and II we can show that the x and y 
in [5] are variables and stand for the coordinates of every 
point on the line MN. Therefore [5] is the equation of MN. 

by § 39. 



40 ANALYTIC GEOMETRY 

EXAMPLES 

I . Which of the four angles made by the axes is crossed by 
the line 

2 3 

Multiplying the equation through by — i we get 

2 3 

Changing two of the signs of the first fraction we get 

—2 3 
This is the form of [5]. Hence 

a^= — 2 ^ the X intercept of the line, 
b ^= 3 ^ the Y intercept of the line. 

X V 

Hence the line — - — ^^- = — i crosses the second angle. 
23 

Which of the four angles made by the axes does each of 
the following lines cross ? 

2. 



3. 
4. 
5. 



PROPOSITION VI 

65. The equation of a straight line in terms of the perpendic- 
ular draw7i to it from the origin^ and the inclination 0/ this per- 
pendicular is 

X cos oL-\- y sin ol — p = o 

in which p is the perpendicular and a the a7igle which it makes 
with the X axis. 



T+ 


y _ 
3 


X 


y _ 


5 


2 


X 


y _ 


3 


^ 


X 

2 


y =2 

3 



THE STRAIGHT LINE 



41 




Fig. 21 

Let MN be any straight line. Let OR be a straight line 
drawn from the origin perpendicular to MN. 

Let /> ^ OR and « = ZiROA. 

Let P be any point on the line MN and draw its coordi- 
nates PK and OK. 

Let ;tr = OKandjK = PK. 

We are to prove that the equation of MN is 

X cos oc-^ y sin a — p = o. 

Draw PH || OX. 

The triangles PBH, ORB and ORA are similar. 

by Geom. 54. 
[i] Hence ^OBR = Z.'ROA = a byGeom.31. 

[2] and ;^ : OB : . PH : PB, byGeom.31. 

[3] hence /.PB = OB.PH= (BH+j/);r = BH..r-(-;t:;i/. 

BH 



[4] Now PB = 

[5] hence by [i] PB = 

[6] and 



cos OBR' 
BH 



by Trig. 2. 



cos a 
x= PH = BH.tan a. 



by Trig. 3 

Substituting the values of PB and x found in [5] and [6] 
into [3] we get 

BH 



[7] 



P 



cos a 



BYL.x + y.BH.tan a, 



42 ANAL-YTIC GEOMETRY 

ro-i P I sin a , >xA . ^ 

[8] or —t^~ =x-\-y ■, by Trig. 6. 

-* cos ^ -^ cos «: ^ o 



[9] hence ;i; cos «: +jk sin «: — p=^o. 



Q.E.D. 



66. Corollary I. — The equation of a straight line i7i terms of 
the perpendicular drawn to it from the origin^ and the angles 
which this perpendicular snakes with both axes is 

X cos a -\-j/ cos /3 — p = o. 

Forlet y^^ZlBOR. 
[i] Then cos /? = sin or. by Trig. 20. 

Substituting this value of sin a into [9] we get 

[2] ^ cos «r -|- J/ cos /? — p = o 

EXAMPLES 

1. YV^3-^~l"ij^^5is the equation of a line in terms of the 
perpendicular drawn to it from the origin and the inclination of 
the perpendicular. 

What angles does the perpendicular make with the axes ? 
What is the length of the perpendicular ? Ans. a = 30° 

/? = 60^ 

P = 5 

2. A line drawn from the origin perpendicular to a second 
line is 3 inches long, and its inclination is 45°. What is the 
equation of the second line in terms of the perpendicular to it 
from the origin, and the inclination of the perpendicular ? 

3. A line whose inclination is 150° cuts y =sx five inches 
from the origin. What is its equation in terms of the perpen- 
dicular to it from the origin and the inclination of the perpen- 
dicular ? 

PROBLEM 

To prove that 

2X V 

[i] 2 I J = f {3^~\-J) — 5 is the equation of 

a straight line. 



THE STRAIGHT LINE 43 

Clearing the equation of fractions we get 

[2] 24 — %x-\-\y — gj/= 24.x -\- Sy — 60. 

Transposing we get 

[3] —137=32-^ — 84. 

Dividing by — 13 we g«t 

[4] ^ = -fl^ + H. 

Now [4] is of the form j = const, x -f" const. Hence it is 
the equation of a straight line. by. § 54. 

PROPOSITION VII 

67. Every equation of the first degree contai7iing two variables 
only is the equation of a straight line. 

For, as in the preceding problem, by clearing of fractions, 
transposing, etc., every such equation can be reduced to the 
form 

[i] Aj/=B;t:+C, 

in which A, B and C are constants. 

Dividing [i] through b}^ A we get 

r 1 B , C 

1 . -, B , C 
m which —^ and — r- are constants. 
A A 

Equation [2] is of the formj/= const. ^ + const. 

But every equation of that form is the equation of a straight 
line. by § 54. 

Therefore every equation of the first degree containing 
two variables only is the equation of a straight line. 

Q. E. D. 

EXAMPI^KS 
I. What are the slope and the intercepts of the line 

■^+2 . y X / s\ I :> 

4 3 ^ 



s 



44 ANALYTIC GEOMETRY 

Clearing the equation of fractions, transposing, etc., we get 
[i] 3j;+52y=66, 



L2] 



j^ = — A-^ + fi 



Hence 



52 



the slope, 



f f = the Y intercept, 
22 =: the X intercept, 



by § 53- 
by § 53- 
by § 45- 



What are the slope and the intercepts of each of the fol- 
lowing: lines ? Draw the lines. 



2. 



X 

3 



2(j^— 3) 



= \y 



2X 



+-I. 



2{x-\- y)~ 



X 



'■y — z-\- 



2X 



4- 



-(t-3-^) = *K— ) + - 



PROPOSITION VIII 
68. If the equation of a straight line be given in the form 

Ax + By+C=zo, 



then by dividing this equation through by — ]/ A^ •\' B"^ ^ we will 
get the equation of the same line in the form 

X cos a -{- y sin ex — p i=: o. 

Y 




Fig. 22 



THE STRAIGHT LINE 45 

In Fig. 22 let 

[i] Ax + Bj'+C^o, 

be the equation of the line AB. From O draw OD_LAB, 
and let 



We are to prove that if we divide [i] through by — y K^-\-W 
we will get the equation of AB in the form 



Dividing [i] through by — V ^ + ^^ we get 

r -, A B C 

[2] >x y =0. 

l/A^+B^ l/A^ + B^ |/A^ + B^ 

[3] Now 



A AB'^ 



l/A^+B^ 



Xb^/a^+b^ 



B B 



^' -A^ VU) +(a) 

From [i] we get 
r 1 AC 

Making ;i; = o in [4] we get 

[5] --^=0K. by §46. 

Making j^ = o in [4] we get 

[6] -^ = 0H. by§45- 

C C 

Substituting these values of ^^ and r- into the last 

B A 

member of [3] we get 

r -, A _ OK _ OK 

L7 J — , ^ ^^ — -^^^. by Geom. 26. 

|./A^ + B^ l/OK-^ + OH^ KH 



46 ANAL YTIC GEO ME TR Y 

OK 
[8] But -=zr^ = sin (p = cos a, by Trig, i and 20. 

Jvxi 

[9] hence — - = cos a. 

^ B 



B AB 

[10] Similarly 



vA' + B^ _C 



XbvA-^+b^ 



A _ OH 

KH- 



x/(l)+(f)' 

|_i[J But rr^7^ = cos ^ = sm <a:. 

B 
[12] Hence — = sm a. 

yA' + B' 

C C B.C. 

by [12]. 
But by [5] 

[14] - ^ = OK. 

Hence [13] becomes 

C 

[15] - = OK sin a. 

l/A-' + B^ 
But by Trig, i 

[16] OD = / = OK sin OKD -= OK sin a. 

by Geom. 54 and 31. 

Hence [15] becomes 

[17] =A 



THE STRAIGHT LINE 47 

Hence by [9], [12J and [17], [2] becomes 

[18] ;ir cos o' + _>' sin a: — p^zo. 

Q. E.D. 

6g. Scholium. — Let us agree to call the direction from the 
origin to the line AB the positive direction. Then p will 
always be positive. Let us agree also to use the positive 
value of -|/A^ + B"''. Then from [17] we see that if, in any 
equation of the form 

Xx + By + C — 0, 

the C be negative, we must change the signs of all the terms 
in the equation, so as to make C positive, before computing 
the value oi p. 

EXAMPLES 

[i] Given the equation of a line in the form 

3^+4j+ 15 = 0, 
to find the equation of the same line in the form [18]. 

What are the values of cos a^ sin a^ and/ ? Draw the line. 

Ans. — f^ — 4)/ — 3 =:= o. 
cos a zzz — |-. 
sin a ^ — i. 

/ = 3. 

[2] Given the equation ^x — 3/ — 15 = o, to change it into 

the form [18]. Ans. fjr — fy — 3=0. 

[3] Given the line 6y — 8-r = 5, to change it into the form 

[18]. Ans. — t'o^ + t'oJ— 2 =0. 

[4] Find the length of the perpendicular drawn from the 

origin to the line i2y — 5^1; = 26, and the inclination of this 

perpendicular. Ans./ = 2. 

Inclination = 22° 37'. 

PS.OPOSITION IX 

70. The distance from any point whose coordiitates are x' and 
y' to the line x cos a -\- y si?i cc — p =. o, is 

x' cos a -\- y' sin oc — /. 



48 



ANALYTIC GEOMETRY 




Fig. 23 

Let M be any point, and AB any straight line. 

Let X cos oL-\- y sin a — p ^ o be the equation of AB. 

Draw ML _L AB. Let D = ML. 

Let OH and MH be the coordinates of the point M, and let 
-;c'=OH,y=MH. 

We are to prove that 

[i] D = ;f ' cos <a^ -f- j^' sin «: — p. 

Draw OK _L AB. 
Then since x cos (x-\-_y sin a — /> = o is the equation of AB, 



[2] 


a /KOA, 


by §65. 


[3] and 


p — OK. 


by §65. 


Through M draw 


A'B' II AB. 




Then 


A'B' 1 OK. 


# 

by Geom. 6. 


Let 


/—OK'. 




Then the equation 


of A'B' is 




[4] X cos 


i a -\- J/ sin a — p' = 0. 


by § 65. 


[5] Hence 


/>'—/= KK'. 




[6] But 


KK'=LM=D, 


by Geom, 17. 


[7] hence 


D =/' —p. 




[8] Hence 


P' = P + D. 





Substituting this value of /' into [4] we get 



THE STRAIGHT LINE 49 

[9] X cos oi +JJ/ sin a — p — D = o, 

for the equation of A'B'. 

Now, since the point M is on the line A'B', its coordinates 
x^ and J/' must satisfy the equation of A'B'. 

Hence substituting jr' andjj/' into [9] we get 
[10] x^ cos Oi -\-y sin a — p — D = o. 

[11] Hence T) =^ x' cos o^ -\- y' sin a — p. 

Q. E. D. 

Now by §69 the direction from O to K is the positive direc- 
tion. Hence from [5] and [6] we see that D is positive. 
Hence we have the following corollary : 

7 1 , Corollary i . — When D is positive , the point from which the 
perpendicular is drawn is on the opposite side of AB from the 
origin ; when D is 7iegative, the point from which the perpen- 
dicular IS drawn is on the same side of AB as the origin. 

T2. Corollary 2 . — Or, when D is positive , the direction from the 
line AB to the point from which the perpendicular is draw?i, is 
the same as the direction from the origin to the li?ie AB ; and 
when D is negative, the direction from AB to the point from 
which the perpendicular is drawn is opposite to the direction from, 
the origin to AB. 

73. Corollary ^. — The distance from the point whose coordi- 
nates are x\ y\ to the line Ax -f- By -[- C= o is 

Ax' + By+C 



D=db 



VK' + B^ 



For by [9], [12] and [17] of § 68 

A 



cos <a^ = — 



sm Of 1= — 



l/A^ 


'+B' 




B 


Va' 


+ B^ 


c 








50 ANALYTIC GEOMETRY 

Substituting these values of cos a^ sin a^ p into [ii] we get 

Vl/A'+B" t/a^' + b' i/a^'+bv' 
Ax' + B j>' + C 

But the direction in which D is drawn from the line AB 
may be the same as the direction of p, or the opposite. 
Hence D may be positive or negative. Therefore [13] may 
be written 

P^^ Ax^+By + C 
V kJ + B'^ 



EXAMPLES 

1. Find the distance from the point 3, 2 to the line 

3XH-4JJ/ — 5=0. 

Draw a figure showing the line, point, and distance. 

Ans. D = 2f. 

2. Find the distance from the point — 2, 5 to the line 

4-^— 3J+8=o. 

Draw a figure showing the line, point, and distance. 

Ans. D — — 3. 

3. Find the distance from the point — 3, — 3 to the line 

3j/ + 1 1 = — 4:r. 

Draw a figure showing the line, point, and distance. 

Ans. D = — 2. 

4. Find the distance from the point i, — i to the line 

'^x — 12 = 4JK. 
Draw a figure showing the line, point, and distance. 

Ans. D — — I. 



CHAPTER VI 

Oblique Axes 

74. Hitherto we have drawn the axes of coordinates per- 
pendicular to each other. It is often more convenient, how- 
ever, to take for the axes two lines which are oblique to each 
other. 

When the axes are perpendicular to each other they are 
called Rectangular Axes, and the coordinates of a point are 
called Rectangular Coordinates. 

When the axes are oblique to each other they are called 
Oblique Axes, and the coordinates of a point are called Ob- 
lique Coordinates. 

In both cases the axes are called Rectilinear Axes and the 
coordinates of a point are called Rectilinear Coordinates. 



PROPOSITION X 
75. When the axes are oblique the equation oj a straight line is 

sin I , , 

-^ stn (go — /) 

in which b is the Y'intercept, I the inclination of the line^ and go 
the angle between the axes. 



Y 




52 ANALYTIC GEOMETRY 

lyet YY' be the Y axis, and XX' the X axis, 
and c» = ^YOX. 

I^et PM be any line and let P be any point on that line. 

Draw PK || YY'. 

Let X = OK, y = PK, d = OL, and /=z: LMO. 

We are to prove that the equation of PM is 

sin / , , 

JK = 7 77 X-\- d. 

-^ sm (go — /) 

Draw OH parallel to PM. 

r n ^, HK sin HOK , ^ . 

I I Then ;— -— = — — ^^^ -. bv Trisr. 14. 

^ -^ OK sm OHK - ^ ^ 

[2] But ^OHK = z:LOH, byGeom. 7. 

[3] and z:loh = z:i.ok— z:hok. 

[4] Hence Z.OnK = Z.LOK — ^liOK. 

Substituting this value of z^OHK into [i] we get 

HK_ sin HOK 

^^^ OK "~ sin (LOK — HOK)' 

[6] z: HOK r= z£ LMO = /, byGeom. 8. 

[7] and ZlIyOK— z:HOK= r«9 — /. 

HK sin / 



[8] Hence 



OK sin ((» — /)' 



[9] and HK = . ^^" ^ ,, OK. 

sm (Gi9 — /) 

[10] Now PH = OL. by Geom. 17. 

Adding [9] and [10] we get 

[11] PKzrz . f^^ OK+OIv, 

sm [00 — /) 
[12] or y = - — 7 7^^-r^- 

/- -• -^ sm (Ce? — I) 

Q. B. D. 

76. Corollary I . — : The equation of a straight li?ie through the 

. . . sin I 

oris:in is y = — — ; — - — ^r- x. 
^ -^ stn {go — I) 



OBLIQUE AXES 53 

76^. Corollary 2. — The equation of a line parallel to the Y 
axis is X ^^ c^:^ any const, 

77 . Scholium. — If the axes be made rectangular, then od = 90° 
and 

sin / sin / sin / 

sin {oD — /) sin (90° — /) cos/ 

But by § 52 tan /is the slope of the line PM, which, as in 
§ 53. we represent by s. Hence when the axes are rectangular 

sin / 
sin (gj — /) ' 
and [12] becomes 

y~sx+d, 
as in § 53. 

PROPOSITION XI 

78. Whe7t the axes are oblique the equation of a straight line 
passing through a fixed point is 

, sin / 

•^ szn (go — /) 

in which x' and y' are the coordinates of the fixed pointy I the in- 
clination of the line and go the angle between the axes. 




Fig-. 25 



Let P' be any fixed point and MN any straight line passing 
through that point. 



54 



ANALYTIC GEOMETRY 



Let P be any point on MN and draw PH and P'H' H YY' 
and PK tl XX'. 

Let ;r = OH and j^ = PH, 
^'=OH' " y=P'H', 
and/=PRH" g^= YOX. 

We are to prove that the equation of MN is 

sin / 



y —y — 



(x' — x) 



[I 

[2 

[3 
[4 

[5 

[6 

[7 
[8 

[9 



sin {00— I) 

Z. P'PK = Z PRH = /. by Geom. 8 . 

Z:RL0 =^YOX — z^LRO. by Geom. 13. 
But Z. RIvO = Z PP'K. by Geom. 8. 

Hence Zl PP'K^^Z! YOX — z:LRO== &?— /. 

P'K sin P'PK sin/ 



hence 
But 



PK sinPP'K sin ((i? — /)' 
sin / 



by Trig. 14. 



P'K = ^ 



PK. 



sin (&7 — /) 
P'K= P'H'— PH —y—y, 
and PK = OH' — OH = Jc'— -r. by Geom. 17. 

sin / 



Hence J/' — y = 



sin (Cfi? — /) 



{x' — x). 



by [6]. 

Q. B. D. 



79. Scholium. — If the axes be made rectangular, then 
CD = 90° and 

sin / sin / sin / 



sin (C89 — /) sin (90° — /) cos/ 



= tan /. 



But tan /is the slope of the line MN, which, as in § 53, we 
will represent by 5. Hence when the axes are rectangular 

sin / 



sin {oa — /) 



s. 



and [9] becomes 

as in § 57. 



y — y = ^ {^x^ — x^ 



OBLIQUE AXES 55 

PROPOSITION XII 

80. When the axes are oblique the equation oj a straight line 
passing through two given points is 



y —y — ZT, — ~, (^' —^) . 



in which x' and y are the coordinates of one oj the given points 
and x" andj/" are the coordinates of the other. 




Let P' and P" be the two given points, and MN a line pass- 
ing through them. 

Let P be any other point on the line MN and draw the 
coordinates of P, P' and P". 

Let;i: = OH and 7 = PH, 
x' =0H' *' y =P'H', 
^" = 0H" " y =P"H". 

Draw PK and P'K' |1 XX'. 
The demonstration is the same as in § 58. 

EXERCISES 

1. Prove that the lines jj/= -^ + i, J^= 2.^+ 2,jk = 3^ + 3 
intersect in the point ( — 1,0). 

2. Find the angle which the lines yc — y = 5, 2y — x ;= 8 
make with each other. Ans. 45°. 



56 ANAL YTIC GEO ME TR Y 

3. Find the equations of the lines passing through the point 
(3, — i) and making angles of 45° with 2/ + 3-^ = 6. 

Ans. 5jK + ^ + 2 = o. 
y — ^x-\- 16 = 0. 

4. Find the equation of the line passing through ( — 3, 5) 
perpendicular to 3jj/ — 2x — 2 = 0. Ans. 2y-\-'^x — i =: o. 

5. Theverticesof atriangleare (2, 3), (4, — 5), ( — 3^—6). 
What are the equations of its sides ? 

Ans. X — 7JK = 39. 
^x — ^y—^. 
4.x -\- y =1 II. 

6. Let the sides of a triangle be given by jj/ = ji: -|- i , jr == 4, 
jj/ = — X — I . What are the equations of the sides of the tri- 
angle formed by lines joining the middle points of the sides of 
the given triangle ? Ans. jk = — .r + 4. 

y — x — 4. 
2X—2,. 

7. Required the equation of a straight line passing through 
the origin and the intersection of the lines 3-r — 2jj/ + 4 = o 
and 3x + 4_>^ := 5. Also find the distance between the two 
points. Ans. gx + 2y = o. 

8. A parallelogram is formed by the lines 

X ^=^ a ] x^=-b\y^^c', y^d\ 

what are the equations of its diagonals ? 

Ans. {d — c)x — {b — a)y:^ad — be. 
{d — c)x-\- (b — a)y ^= bd — ac. 

9. Find the value of ^ provided the line y ^ sx passes 
through (i, 4). Ans. ^=4. 

10. Required the equation of a straight line perpendicular 
to Ax + B/+ C = o and making an intercept b on the Y axis. 

Ans. Bx = A{y — b). 

11. What is the equation of the line perpendicular to 



X . y 

Ans. ax — by = a^ — b 



+ ^ = I and passing through {a, b) ? 



OBLIQUE AXES 57 

12. Required the equations of the perpendiculars erected at 
the middle points of the sides of the triangle whose vertices 
are (5, — 7), (i, ii), ( — 4, 13). Prove that these perpen- 
diculars meet in a point. 

13. The points (i, 2), (2, 3) being equidistant from the 
point {x, y) , write the equation which expresses that fact. 

Ans. {x-ir-^{y~2Y = {x-2y+(y-3y\ 

or jf-f-jK = 4. 

14. Find the distance between the points (x" , j/") , {x.',y), 
the axes making the angle cp with each other. 

Ans. D=i/{x"—x'y+{y"—yy+2{x"—x'){y'—y)coscp. 

15. What is the length of the perpendicular from the origin 

to the line h ^ = i ? 

3 4 

16. Find the point equidistant from the three lines 

4^ + 3_>/ — 7 = 0, 5^+ i2_y — 20 = 0, 3^ + 4jj/ — 8 = 0, 
and its distance from each. Ans. (2,3); 2. 

17. Two lines are given, one passing through the points 
(i, 2), ( — 4, — 3), the other through (i, — 3), and making 
an angle of 45° with the first ; what are the equations of the 
lines? Ans. ^ = x-|- I and^ = — 3. 



CHAPTER VII 

Transformation to New Axes 

PROPOSITION XIII 

81. If we have given the equation of a locus referred to any 
pair of axes ^ we ca7i find the equation of the same locus referred 
to any pair of axes parallel to the first by putting 

m -\- x' for X, 
and n +jk' for y, 
'in the original equation ; m and n being the coordinates of the 
new origin^ and x^ and y the coordi7tates of any point on the locus 
referred to the new axes. 




In Fig. 27 let OY and OX be the original axes, and O'Y' 
and O'X' be the new axes. 

Let RS be any locus and P an}^ point on it. 

Draw PH parallel to OY. Then OH and PH will be the 
coordinates of the point P referred to the original axes, and 
OX and PI< will be its coordinates referred to the new axes. 

Ivet m = OK and n = O'K, 
x'^=OX "y = PL. 



TRANSFORMATION TO NEW AXES 59 

[i] Let f{x,y)—o, 

be the equation of RS referred to the original axes. Then 

;t = OHandjK= PH. 

We are to prove that we can find the equation of RS re- 
ferred to O'X' and O'Y' by putting 

ni-\- x^ for X 

and ^ +JK' for y 
into [i]. 

[2] Now OH = OK + OX, by Geom. 17. 

[3] or X ^^ 7n -\- x\ 

[4] and PH = PL + O'K. by Geom. 17. 

[5] Hence y ^^ n -\;- y . 

If we substitute these values of x and jj/ into [i] we get 

[6] f{m + x\ n +y) = o. 

In this equation m and n are constants, but x^ and y' are the 
coordinates of any point on RS. Hence [6] is the equation 
of RS referred to the new axes. by § 39. 

Q. E.D. 

PROPOSITION XIV 

82. If we have given the equation of any locus referred to any 
pair of axes, we can find the equation of the same locus referred 
to any other pair of axes having the. same origin by putting 

x^ sin (6 — a) + y' sin (6 — 6) . 

—^ for X, 

sin u 

^ x' sin oc-{- y' sin (3 

and -. — =k for y, 

sin u 

into the equation of the locus ; x' and y' being the coordi?iates of 
any point on the locus referred to the new axes, 6 the a?igle be- 
tween the original axes, a the inclination of the new axis of X, 
and § the inclination of the new axis of Y. 



6o 



ANALYTIC GEOMETRY 




In Fig. 28 let RS be any curve and P any point on it. 

Let OY and OX be the original axes and OY' and OX' the 
new axes. 

Draw PM || OY and PN || OY'. Then OM and PM are the 
coordinates of P referred to the original axes, and ON and 
PN its coordinates referred to the new axes. 

Let ^' = ON andy = PN. 

[i] Let f{x,y) = o, 

be the equation of RS referred to the original axes. Then 

j»r=OM and ^'^PM. 

Let a = X'OX, 
yS = Y'OX, 
8 = VOX. 

We are to prove that we can find the equation of RS re- 
ferred to OX' and OY' by putting 



and 
into [i]. 



x' sin {6 — a) +/ sin {0-/3) 
sin 6 

x' sin (X -\- y' sin /^ 



sm 



e 



for X, 
for y. 



TRANSFORMATION TO NEW AXES 6i 

Draw any line OL through the origin and let 

^ = z::ivOX. 

Draw PH, NJ and MK _L OL. 
Draw NT and MQ Ij OL. 

[2] Now OK = OM cos q). by Trig. 2. 

[3] Hence OK = x cos 99 ; 

[4] also MQ= PM cos PMQ, by Trig. 2. 

[5] but MQ = KH, byGeom. 17. 

[6] and z^PMQ =^YOH= ^ — 0. byGeom. 11. 

Substituting these values of MQ and PMQ into [4] we get 

[7] KH=_)/cos (cp—d). 

Adding [3] and [7], member to member, we get 

[8> OK+KH:=^ cos cp+y cos (cp—O), 

[9] or OH = Jt: cos ^ +JJ/ cos (^ — 6). 

[10] Similarly OJ = ONcos(t^ — ry) =: x' cos (cp — ex), 
[11] and JH = NT= PN cos PNT = >/' cos {cp — (3) . 

Adding [10] and [11], member to member, we get 

[12] OH = x' cos {cp — a) -\-y cos {'P — ft). 

Hence from [9] and [12] we get 

[13] 

X cos (p -\- y cos {cp — 6) ^ x^ cos {cp — ol) -\- y cos{<p — ft). 

Since OL is any line drawn through O, [13] must be true 
whatever be the angle (p. Hence we may draw OL so that 



[14] 


9= l+e. 


see § 42, Cor 


[15] then 


7t 
2 




[16] Hence 


COS cp = — sin Q, 


by Trig. 21. 



62 ANALYTIC GEOMETRY 

7t 

[17] and cos (<7? — 0) = cos — =0. by Trig. 19. 

Again by [14] 

n 



[18] cos {cp — a) = cos ( \-0 — a) = — sm(0 — a). 

by Trig. 21. 
[18'] Similarly cos {(p — /3) = — sin {0-/3). 
Hence by [16], [17], [18] and [18'], [13] becomes 
— .r sin d =: — x' sin {d — a) — y' sin {6 — /3) . 
jr'sin {d—a) +ysin (0 — /3) 



[19] x = 



sm 



d 



Again, since [13] must be true whatever be the angle (p, 
we may draw 01^ so that 

^ = — . see § 42, Cor. 

Then [13] becomes 

y sin = x' sin (x -\- y' sin /?, by Trig. 19 and 20. 

r- -, , x' sin a -4- y sin /3 

I 20 and y= -. — ^ . 

sm C7 

Substituting the values of x andy found in [19] and [20] 
into [i] we get 

[21] 

'x' sin {6 — a) +jj/' sin {8 — /3) x' sin a -j-y' sin /S' 



A 



o. 



sin 6 ' sin 

Now in [21] a^ /? and are constants, but x' and j/' are the 
coordinates of any point on RS. Hence [21] is the equation 
of RS referred to the new axes. by § 39. 

Q. E. D. 

83. Cor. — If we have given the equation of any locus referred 
to any pair of rectangular axes, we can find the equation of the 
satne locus referred to any other pair of rectangular axes having 
the same origin by ptitting 

x' cos a -\- y si7i oi for x, 



TRANSFORMATION TO NEW AXES 63 

and x' sin o^ -{- y' cos a for y^ 

into the equation of the locus. 

For, if the axes are rectangular, then in Fig. 28 

[22] e = — , 

[23] Hence -^— ^YOX' = a:. 

[24] Also z:y'ox'=-^. 

2 
[25] Hence ^ ~ Z.^01^ = ZM'ON , 

[26] and by [23] Z. Y'OY = a. 
Now by Fig. 28 

/j = o + z:y'oy. 

[27] Henceby [22] and [26] /^ = 1- <af. 

Hence since C7 = — , 

2 

and p = — +^> 

2 

equations [19] and [20] become 

[28] jc = ;t:' cos <a' — y sin <x, 

[29] jj/ = ;r' sin ^ +j' cos oc. 

Substituting these values of x andjF into [i], we get 
[30] f{x^ cos OL — y^ sin a^ x' sin a -|-ji^'cos«^) = o. 

Since x' and jf' are the coordinates of any point on RS, [30] 
must be the equation of RS referred to the new axes. 

Q. E.D. 



64 ANAL YTIC GEO ME TR Y 

PROPOSITION XV 

84. If we have given the equation of a curve referred to any 
pair of axes, we may find the equation of the same curve referred 
to any other pair of axes by putting 

. x''si?i(qj — a) -\- y" sin (o) — /3) 

m -\ -^ 7^-^ ^-^ -— ^ for X, 

sin (p 

. ' . x" sin ex 4- y" si7i . 

a?ia n-\ -. =^- for 1/, 

siti cp 

into the equation of the curve ; m and n being the coordinates of 
the new origin^ arid x" and y" the coordinates of any point on the 
curve referred to the new axes, a and ft the inclinations of the 
new axis of X and axis of Y respectively to the original axis of 
X, ajid cp the angle between the original axes. 

ft 



P 



8/X" 




Fig. 29 

In Fig. 29 let OX and OY be the original axes, and let 
O'X" and O'Y" be the new axes. 

Let OK and PK be the coordinates of P referred to the 
original axes. 

[i] Let f{x,y)=o 

be the equation of any curve RS referred to the original axes, 
and P be any point on that curve. 

Then ji: = OK andji/ = PK. 

Let O'H and PH be the coordinates of P referred to the 
new axes, and let x"^: O'H andj/"= PH 



TRANSFORMATION TO NEW AXES 65 

The coordinates of the new origin are OQ and O'Q. 
lyet ;;? ^ OQ and n = O'Q. 

/?=z:y"vx, 

cp-^^Z. YOX. 

We are to prove that the equation of RS referred to the 
new axes is 

^" sin ( cp — a) + jj/" sin ( ^ — ^) 



f\m + 



sm q) 



. .r"sin «'+J^"sin^\ 

^ )=o. 

sm <z> / 



n 

9 



Through O' draw O'X' || OX and O'Y' |! OY. The coordi- 
nates of P referred to O'X' and O'Y' will be 

;tr' = 0'J andy= PJ. 
[2] Then x—m-{-x\ by § 81. 

[3] and y^=^n-\-y\ by § 81, 

[4] But^' = ^"^^°^^~">.+-^"^'"^^-^\ by §82. 
L^-" sm cp ^ 

r -. 1 / ;r"sin ar+J^"sin/5) , ^^ 

[5] and y— ■■ ^:^^ ^-^. by §82. 

^^■J -^ sm cp ■' 

Substituting these values of x^ and y into [2] and [3] we 
get 

;r"sin {cp — «)+y' sin {cp — yS) 



[6] x^=-7n-\ 



[7] y = n-^ 



sm cp 
;i;"sin a-\-j/" sin /? 



sm (p. 

Substituting these values of x and jv into [i] we get 
x" sin (cp — a) -{- y" sin (cp — /?) 



[8] f(m + 



sm cp 

, x" sin oi 4- y" sin /3\ 

n -\ -. — I = o. 

sm cp J 

Now in [8] the ;«:" and y" are the coordinates of any point 
on the curve RS referred to the new axes, but all the other 



66 ANALYTIC GEOMETRY 

quantities in the equation are constants. Hence [8] is the 
equation of RS referred to the new axes. by Trig. 39. 

EXAMPLES 

1 . Required the equation of the line j/ = 3;^ + i when we 
remove the origin to (2, 3), the axes remaining parallel to 
themselves. Ans. y =^ -^x -\- \. 

2 . Given the equation j/^ + ^^ — 4-^ + 4 JJ^ — 8^0; required 
its form when the origin is at (2, — 2) Ans. x^ +y^ ^ 16. 

3. If we turn the axes through an angle = 45°, what does 
the equation jr' +j^^ = a^ become ? 

4. What does the equation x'^ -\- y^ — \x — 6ji/ = 18 become 
when the origin is at the point (2, 3), the axes still being rec- 
tangular ? 

5. What do the following equations become when the ori- 
gin is changed to the point given ? 

X -\- y -(-2 = 0; the new origin ( — 2, o) . Ans. x -\- y ^o. 
3x' + 4xy + y' — sx — 6y — 3 = 0; (|-, —4). 

Ans. i2jr^ + i6xy-\- 4jj^^= i. 
x^ — 6xy -\- y^ — 6x -\- 2y -\- i ^ o; (o, i ) . 

6. Show that the degree of an equation is not altered by a 
transformation of coordinates. 



CHAPTER XVIII 

Polar Coordinates 

85. The position of a point may be indicated by giving its 
distance and direction from a given fixed point in a given 
fixed straight line. 

H 




Fig. 30 

For instance, if OX be a given fixed straight line, and O a 
given fixed point in it, then we may indicate the position of 
any point P by giving the angle, which a line drawn from O 
through P makes with OX, and also the distance along this 
line from O to P. 

86. The Initial Line. — The given fixed straight line is 
called the initial line. 

87. The Pole. — The given fixed point is called ^^pole. 

88. The Radius Vector. — A straight line drawn from the 
pole to any point is called the radius vector of that point. 

OP is the radius vector of the point P. 

89. The Vectorial Angle. — The angle between the initial 
line and the radius vector is called the vectorial angle. 

POX is the vectorial angle. 

go. The Vectorial Arc. — Th.Q. vectorial angle is measured 
by an arc whose center is the pole, and which lies between 



68 ANALYTIC GEOMETRY 

the initial line and the radius vector. This is called the vec- 
torial arc. 

AB is the vectorial arc. 

91. The z^^<:/^rz«/ «r<: is considered positive when it is gen- 
erated by a point revolving about the pole, from the initial 
line, counter clockwise. 

92. The vectorial atigle is considered negative when it is 
generated by a point revolving about the pole, from the ini- 
tial line, clockwise. 

93. The vectorial angle is said to be positive when the vec- 
torial arc is positive, and negative when the vectorial arc is 
negative. 

94. The radius vector is positive when it is drawn from the 
pole towards the terminus of the vectorial arc. 

95. The radius vector is negative when it is drawn from the 
pole in the direction opposite to that of the terminus of the 
vectorial arc. 

In Fig. 30 OP is positive. 

96. We may indicate the position of the point P' by giving 
the angle BOA and the distance OP'. In that case OP' is 
negative. 

The radius vector is denoted by r. 
The vectorial angle is denoted by ^. 

EXAMPIvES 

Locate the points whose coordinates are given in the follow- 
ing examples : 

I. ^ = — 5>^ = 45°- 




POLAR COORDINATES 
2. r— 5, (9= 225°. 



69 




3. r— 10, Q— 135°. 

4. r— — io,d— I35^ 

What is the position of the pole with respect to the two 
points in Examples 3 and 4 ? 

5. r— 10, Q — 315°. 

What is the position of this point with respect to that in 
Example 4 ? 



PROPOSITION XVI 
97 . The polar equation of a straight line is 



P 



COS {6 — ^)' 

in which p represents the perpendicular from the pole to the line, 
a the inclination of this perpendicular, p the radius vector of any 
point on the line, and Q the vectorial angle. 




Fig. 31 
I^et OX be the initial line and O the pole. 



70 ANALYTIC GEOMETRY 

Let MN be any straight line, OK the perpendicular drawn 
from the pole to this line, and P any point on the line MN. 

Let «^ = KOL and;^ = OK, 
<9 = P0X " p = OP. 

We are to prove that 







H — 

cos 


{d-ay 




[I] 




cos POK = ^, 


by Trig. 2. 


[2] 


or 


cos (^ — 


p 




[3] 


Hence 


f\ 


p 




r — 

COS 


{d-ay 





Q. E. D. 

98. Corollary /. — The equation of a li?ie perpendicular to 
the initial line is 

._ P 



cos 



0' 



For if MN be _Lto OX, OK must coincide with OX. 

by Geom. 5. 
[i] Hence «r = o, 

and [3] will become 

[2] P^-^' 

-* cos u 

99. Corollary 2. — To find where a straight line cuts the initial 
line make 6 z=i o, and then p will be the distance from the pole to 
the cutting point. 

For, since in Fig. 31 P represents any point on the line MN, 
it may represent the point L. 

Then ^ = 0, and by [3], §97 

[i] p= f- .=:-^. by Trig. 12. 

■- -" cos ( — ol) cos ol 

[2] But 0L = — ^sT=-^^- by Trig. 2. 

*- ^ cos KOL cos OL 

[3] Hence p = OL. 



POLAR COORDINATES 71 

EXAMPLES 

1. The perpendicular from the pole to a straight line is 5 
inches long and makes an angle of 60^ with the initial line. 
What is the equation of the line? Where does it cut the ini- 
tial line ? 

Here oc = 60° and p -=^ S- 
Substituting these values into [3] of § 97, we get 

' cos ((9 — 60°)' 
which is the equation of the line. 

To find where the line cuts the initial line : 

I^et 0^0. by § 99. 

Then P= / A o, =^= lo- by Trig. 12. 

cos ( — 60 ) ^ 

2. The perpendicular from the pole upon a straight line is 10 
inches long, and makes an angle of 240° wnth the initial line. 
Where does it cut the initial line ? 

Here a ^ 240° and p = 10. 
Substituting these values into [3] of § 97, we get 

10 

^""cosC^ — 240°)' 
which is the equation of the line. 

To find where it cuts the initial line : 

I^et = o. 

^, 10 10 

Then p = 

and 

3. Show that the equation of transformation from a rec- 
tangular to a polar sj^stem of coordinates, the origin, and 
pole being non-coincident, are 

jr = <2 + r cos {0 -\- cp) ^ 

ji' zz: ^ -|- r sin {6 -{- q)) , 
where the origin is {a, d) , the ^cp is the z^of the initial line 
wnth the X axis, and is the vectorial angle. 



cos ( — 240°) 


COS 240° 


10 

p= 1 =~ 

2 


20. 



CHAPTER IX 

The Ellipse 

roo. The Ellipse. — An Ellipse is the locus of a point mov- 
ing in a plane in such a way that the sum of its distances 
from two fixed points in the plane is constant. 




Fig. 32 



Let F and F' be the two fixed points in the plane. 

Let P be a point moving in this plane in such a way that 
PF+ PF' is constant. 

Then the locus ABA'B' traced out by P is an ellipse. 



PROBI^KM 

loi. To draw an ellipse. 

Let F and F' in Fig. 32 represent two fixed pins. Take an 
inelastic thread longer than the distance FF'. Fasten an end 
of it to each pin. Press the point of a pencil against the 
thread so as to stretch it. Then if, in the figure, P represent 
the point of a pencil, F'PF will represent the thread. 

Now move the .pencil so as always to keep the thread 
stretched. Then since the thread is inelastic and is always 
stretched, the sum of the distances PF and PF', from the pen- 
cil point to the fixed pins, is constant. Hence the locus 
traced out by the pencil point must be an ellipse by § 100. 

Q. E.D. 



THE ELLIPSE 73 

102. Corollary. — It is obvious that the ellipse is a closed curve^ 
and that it will cut the straight line drawn through the two fixed 
points F a7id P in two points as A and A'. 

103. The Foci. — The two fixed points F and F' are called 
the foci. 

104. The Focal Radii. — The distances from the foci to any 
point on the ellipse are called the focal radii of that point. 

FP and F'P are the focal radii of the point P. 

105. The Vertices. — The points in which the ellipse cuts 
the straight line which passes through the foci are called the 
vertices of the ellipse. 

A and A' are vertices. 

106. The Transverse Axis. — The line which joins the 
vertices is called the transverse axis of the ellipse. 

AA' is the transverse axis. 

107. The Center. — The middle of the transverse axis is 
called the center of the ellipse. 

108. The Conjugate Axis. — The conjugate axis is a straight 
line drawn through the center perpendicular to the transverse 
axis and terminated both ways by the ellipse. 

BB' is the conjugate axis. 

PROPOSITION I 

109. The sum of the Jocal radii of any point on an ellipse is 
equal to the transverse axes. 




Fig. 33 



74 ANAL YTIC GEO ME TR Y 

Let P be any point on the ellipse, r and r' its focal radii, 
and AA' its transverse axis. 

We are to prove that 

r' + r=: AA'. 

When the moving point P reaches A 

[i] r'+r=FA + AF. by § loi. 

When the moving point reaches A' 

[2] r'4-r=FA'+FA'. 

[3] Hence F'A + AF = FA' + A'F', 

[4] or FF'+ AF+ AF=FF' + A'F' + F'A'. 

[5] FF' = FF'. 

By subtracting we get 
[6] 2AF=2A'F'. 

[7] Hence AF = A'F'. 

Substituting A'F' for AF in [i] we get 

[8] / + r = F'A H- A'F' = AA'. 

Q. E. D. 

1 10 . Corollary. — The foci of an ellipse are equally distant from 
the center. 

For 

[i] CA:=CA', by §107. 

[2I and AF = A'F', ' by § 109, [7]. 

[3] hence CA — AF = CA' — A'F', 

[4] or CF = CF'. 

PROPOSITION II 

111. The equation of the ellipse is 

ay + ^V = a'b\ 

in which a represents the semi-transverse axis, b the semi- conjugate 
axis, and x and y represent the coordinates of any point 07i the 
ellipse. 

Let ABA'B' be an ellipse, AA' its transverse axis, BB' its 
conjugate axis, and C its center. 

Let « ^ CA and /^ = CB. 




75 



Let P be any point on the ellipse and draw its ordinate PK. 

Let Jt = CK and y = PK. 
We are to prove that 

ay + b'x' = a'b' 
is the equation of the ellipse. 

Let F and F' be the foci, and draw the focal radii FP and 

F'P. 

Let r ^ FP and r' = F'P . 



[I 

[2 

[3 
[4 

[5 

[6 

[7 
[8 

[9 
[to 

[II 



Then 

but 

hence 

and 

Again 
but 
Hence 
and 

Now 



Let c = CF. 

CF' = ^, 

2 2 

r' = FK + P"K , 
FK = ;f — ^, 
r-^{x-cy+y\ 



by § no. 
by Geom. 26. 



r=zy{x — cr + y\ 

r" = F'^+PK, 
F'K = Ji;+^. 



by Geom. 26. 



r' -\- r— AA'= 2a, 



by § 109. 



hence i/(:r — cY -\- y"^ -{- \/ {x -\- cY -\- y"^ = 2a. 



76 ANALYTIC GEOMETRY 



[12] V{x — cY^f = 2a — ^{x-^cf^y\ 



[13] {x—cy-{-y'' = \a'—\a^{x-^cy-\-y''-\-{x-^cY-\ry'' 



[14] x'^ 2CX -\-C^ ^r. 4(2^ 4« y ( JT 4~ ^) "^ 4~ J^'"f*^' + 2^^ +<:^. 



[15] ^ay^{x-\-cy +y =^a' + ^cx. 



[16] a i/^ + f)^4-jF^ ^a^+^^-^- 

[18] a^x^ -\- 2a' ex -\- a'^c^ + a^y^ z= a^ -\- 20^ ex + ^V. 
[19] <2y + (<2' — c^)x'^ ^^d^{a^ — r). 

In [19] X andjF stand for the coordinates of any point on 
ABA'B'. Hence the}^ may stand for the coordinates of the 
point B, which are 

X =^0 and jF = b. 
Substituting these values of x and y into [19], we get 
[20] a^b^:=^a^{a~ — r), 

[21] b' ^=- a^ — c . 

Substituting this value of {d^ — r) into [19], we get 
[22] d'y'-\-b''x''^a'b\ 

Since in [22] the x and y stand for the coordinates of any 
point on the ellipse, that equation must be the equation of the 
ellipse. by § 39. 

Q. K.D. 

112. Corollary. — The semi- transverse axis is equal to the 
distance from the focus to the extrcTnity of the conjugate axis. 



[I] 




b'=a' -c\ 


by § III, [21] 


[2] 


Hence 


a'=b' + c\ 




[3] 


but 


F'B = <5' + c\ 


by Geom. 26 


[4] 


Hence 


a'' = F'B, 




[5] 


and 


a = F'B. 





THE ELLIPSE 



77 



The Circle. — The circle is an ellipse in which the conju- 
gate and transverse axes are equal to each other. 

PROPOSITION III 
113. The equation of the circle is 

x'+y = r\ 
in which r is the radius of the circle. 

Y 




Fig- 35 

Let P be any point on the ellipse and PK its ordinate. 
jt: = CK andjK = PK. 
^? = CA and ^ = CB. 
The equation of the ellipse is 
[i] ay-^b^x'z:^a'b\ 

Let bz=^a, 

[2] then b^ — a\ 

Substituting a^ for U^ in [ i ] , we get 
[3] ay'^ + <2 V ^ a'^. 

[4] Hence >'' -\~ x^ — ^^ 



by § III 



[5] or 

[6] but 

[7] Hence 

[8] hence 



PK+CK = CA, 

PK + CK = CP . 

2 2 

CP = CA, 

CP 1= CA. 



by Geom. 26, 



78 



ANALYTIC GEOMETRY 



Now, if the point P be moved along the ellipse, CP will 
always retain the same value, since it is always equal to CA. 
CP is therefore constant. 

Since P is any point on the ellipse, and CP is constant, the 
ellipse must be a circle. by Geom. i8. 

Since in [4] the x and y stand for the coordinates of any 
point on this circle, that equation must be the equation of the 
circle. by § 39. 

Substituting r for the a in [4], we get 

[9] ^' + y=r^ 

which is the equation of the circle when the center is at the 
origin. 

114. Corollary i , — The circle is called the equilateral ellipse. 

115. Corollary 2. — When the center is not the origin^ the 
equation of the circle is 

{x — my-\- (y — n)^^= r^. 

r 




Fig. 36 

IvCt the new origin O be at any other point than C. 
Let Y'C and X'C be the old axes and YO and XO be the 
new axes. 

Let P be any point on the circumference and let 

x^OKand y^ PK, 
^'^CR " y = PR, 
m = OH '' n = CU. 
and^-^CP. 



THE ELLIPSE 



79 



[i] Then 

is the equation of the circle referred to the old axes. 

by § 113. 
[2] But x' =^ X — mandy=^j — n. by Geom. 17. 

Substituting these values of x' and y' into [i], we get 
[3] {x — ^)^ -{- {y — n)^=zr'. 

Since x and y are the coordinates of any point on the cir- 
cumference of the circle, [3] must be the equation of the cir- 
cle referred to the new axes. ^ by § 39. 

116. The Circumscribed Circle. — If from the center of an 
ellipse as a center with a radius equal to the semi-transverse 
axis of the ellipse, a circle be drawn, the circle is said to be 
circumscribed about the ellipse. 

117. Corresponding Ordinates. — Ordinates drawm from 
the ellipse and from the circle on the same side of the trans- 
verse axis, and meeting the transverse axis at the same point, 
are called corresponding ordinates, and the points on the curves 
from which they are drawn, corresponding points. 

PROPOSITION IV 

118. If a circle be circumscribed about an ellipse, any ordinate 
of the ellipse is to the correspoiiding ordinate of the circle as the 
semi-conjugate axis of the ellipse is to its semi-transverse axis. 



X' 







Y 








\ 


D 


P' 




^ 


f 


B ^ 


^N 


X 


1 

A'f 








V 


^ 




G 
Y' 


( 


^ 



'^ig-37 



Ml 



80 ANALYTIC GEOMETRY 

Let P be any point on the ellipse and draw its ordinate PK. 
Produce PK till it meets the circle at some point P'. 

Let X = CK, jK = PK andy = P'K. 
Let « = CA and ^ = CB. 

We are to prove that 

y \ y :: b : a. 

Since the point P is on the ellipse, its coordinates x and y 
must satisfy the equation of the ellipse. by §40. 

Hence, letting the x and y of that equation stand for the 
coordinates of the point P, we have 

[i] ay-{-b'x'^a'b\ by §111. 

[2] hence y^^=i-^{a^ — x'^). 

Since the point P' is on the circle, its coordinates x and j/' 
must satisfy the equation of the circle. by § 40. 

Hence substituting x and y' for the x and y of that equa- 
tion, we get 

[3] ;r'+y' = r\ by § 113. 

[4] Hence j/'^ = ^ — x'. 

But since, by s 116, r^= a, this equation becomes 

[5] y- = a'-x\ 

Now, dividing the members of [2] by the corresponding 
members of [5], we get 

y'_b' 

^ -" y a 

[7] Hence y \ y \\b\a. 

Q. K. D. 

119. Corollary i . — Any two ordinates of an ellipse are to each 
other as the corresponding ordinates of the circumscribed circle. 
That is 

PK : RH : : P'K:SH. 

120. Corollary 2. — If from the center of the ellipse a circle be 
drawn havirig the conjugate axis as a diameter, the abscissa of 



THE ELLIPSE 



8i 



any poirit on the ellipse will be to the abscissa of the corresponding 
point on the circle as the semi- transverse axis is to the semi-con- 
jugate axis. 

121. Corollary j. — To draw an ellipse when its axes are given . 




Let XX' be the.X axis, and YY' the Y axis. 

I^et <2 = CA and ^ = CB. 

On a stiff ruler mark off a length PQ = ^, and a length 
PR = b. Then place Q on the Y axis, as at Q in Fig. 38, and 
place Rupon the X axis, as at R in the figure. ISJow slide 
R and Q along the axes, and P will trace out an ellipse. 

For, circumscribe a circle about the ellipse. 

Through any point P on the ellipse, draw an ordinate PK 
and produce it till it meets the circle at S. 

Let jr = CK and j/ = PK. 



[I] 


Then PK:SK: :b:a. 


by § 118. 


[2] 


Hence PK : \CS— CK : : 3 : a. 


by Geom. 27. 


[3] 


or y -.y' a'^ — x^ : : b : a. 




[4] 


Hence ay^^b-\/a^ — x^, 


by Geom. 21. 



82 



ANAL YTIC GEO ME TR Y 



[5] hence aY -^ U" x' ^ a' b\ 

which is the equation of the ellipse. by § in. 

Therefore, as the points R and Q move along the axes, P 
traces out an ellipse. by § 41. 



PROPOSITION V 

1 22. The squares of the ordinatesofany two points on an ellipse 
are to each other as the products of the segments which they make 
on the transverse axis. 




L,et P' and P" be any two points on an ellipse, and let P'K 
and P"H be their ordinates. 

Let y ^ CK and / = P'K. 
y = CH '' y' = P"H. 

We are to prove that 

y ' : y^ : : A'K.KA : A'H.HA. 
Let « = AC== A'C. 
[i] Then A'K = « + :tr' and KA = « — jc', 
[2] also A'H = « + ;c" '' YLK — a—x'\ 

Since the point P is on the ellipse, its coordinates x' andy' 
must satisfy the equation of the ellipse. by § 40. 

Substituting x' and y' for x and y in that equation, we get 
[3] ay + d'x" = a'd\ by § 1 1 1 . 



THE ELLIPSE 



Since the point P" is on the ellipse, its coordinates must 
also satisfy the equation of the ellipse. by § 40. 

Substituting x^^ and j/" for the x and y of that equation, 
we get 

[4] ay ^ + ^ v'^ = a^/^^ 

By transposing and factoring, [3] becomes 

[5] ay' = b\a'-x"), 

and [4] becomes 

[6] aY'^b^a" —x'^''). 

Dividing the members of [5] by the corresponding mem- 
bers of [6] , we get 

y^ _ a^—x'^ _ {a-\-x'){a — x') 

[8] Hence y- :/" : : {a-\- x^){a — x') : (a+jt:") (a — ^"), 

[9] or y :ji/"' : : A'K.KA : A'H.HA. 

Q. E.D. 
123. Corollary i . — Ordinates at equal distances from the center 
are equal. 

Let CJ = x"' and LJ =y". 
Let CJ = CH. 

[i] Then ^ : y '^ : : AH. HA' : AJ.JA', by § 122. 

[2] CJ z= CH, 

[3] AC = A'C. by § 107. 

Adding [2] and [3] we get 
• [4] AC + CJ = A'C + CH, 

[5] that is AJ = HA'. 

[6] Again A'C = AC, 

[7] and CJ=CH. 

Subtracting [7] from [6] we get 

[8] A'C— CJ = AC — CH, 

[9] that is JA'=AH. 

Multiplying the members of [5] by the corresponding mem- 
bers of [9J, we get 



84 ANALYTIC GEOMETRY 

[lo] AJJA'= AH.HA' 

Hence [i] becomes 



[II] 


ym 


: y " : 


: AH.HA' : AH.HA' 


[12] 


Hence 




.jj/'" 

yun — I- 


[13] 


Hence 




^m _ ^rff2^ 


[14] 


and 




'y^^y\ 



Q. E. D. 

124. The Parameter. — The parameter of an elHpse is the 
double ordinate which passes through the focus. 

Thus RS is the parameter. 

The parameter is sometimes called the latus rectum. 

125. Corollary 2. — The parameter is a third proportional to 
the transverse and conjugate axes. 

For, since the point R is on the ellipse, its coordinates CF' 
and RF' must satisfy the equation of the ellipse. Hence sub- 
stituting CF' for X and RF' for j' in that equation, we get 

[i] «'R"F' + ^^CF =r a^^^ by §111. 



[2] 


But 


CF'=F'B— 3^ 


by Geom. 27, 


[3] 


and 


F'B=r«. 


by § 112. 


[4] 


Hence 


CF'=a^ b\ 




[5] 


Hence 


«^RF' + /^^(a' b'^ =a'b\ 


by [i]. 


[6] 


or 


^'RF' + ^^^~^ b' ^a'b\ 




[7] 


and 


«2RF' = b\ 




[8] 




• «RF'=^^ 




[9] 


Hence 


2^2RF' = ^b\ 




[10] 


and 


2RF : 2b '. : 2b : 2a. 


by Geom. 56. 


[II] 




2RF'=RS 


by § 123. 


[12] 


hence 


RS : BB' : : BB' : AA'. 





THE ELLIPSE 85 

126. Corollary ^. — The ellipse is symmetrical with respect to 
both axes. 

127. Corollary 4. — If the ordinates of any two points on an 
ellipse be equally distant from the center^ the points will be equally 
dista?it from the adjace7it foci. That is 

128. The Eccentricity. — The eccentricity of an ellipse is 
the quotient of the distance from the center to the focus by the 
semi-transverse axis. 







Let ^ 


— the eccentricity. 


n Fi 


g. 39 let 


c CF 


, and a — CA. 


[i] 


Then 




CF c 
^ CA a' 


[2] 


and 




c=^ ae. 


[3] 


But 




a' b' = c\ 


[4] 


Hence 


a:' b' 
a' 


e ( c V 

~ a' ~\ a ) ~' 


[5] 


and 




b' 
2=1 e-. 

a 



by § III, [21]. 



129. Corollary ^ . — The eccentricity of an ellipse is less than i . 
For in [i] CF is less than CA. 

130. Corollary 6. — The eccentricity of a circle is o. 

For when the ellipse takes the form which we call the circle, 
b^^ a and [4] becomes 

*> 7 2 

2 a~ — o 

a~ a^ ' 

EXAMPLES 

I. The semi-transverse axis of the earth's orbit is about 
93,000,000 miles, and its eccentricity is -^. 

Find the length of the conjugate axis, the distance from the 
focus to the center, and the greatest and least distance of the 
earth from the sun, which is at one of the foci. 



86 ANALYTIC GEOMETRY 

2. The semi-transverse axis of Jupiter's orbit is 483,000,000 
miles, and its semi-conjugate axis is 478,000,000 miles. 

Find the eccentricity of the orbit, the distance from the 
focus to the end of the conjugate axis, and the greatest and 
least distance of Jupiter from the sun, which is at one of the 
foci. 

3. Find the semi-axes and eccentricity of each of the follow- 
ing ellipses : 

25 jj/^ + i6jr' = 400. 

3 2 

rj/^ -|- -^^ = ^' 

4. The equation of an ellipse is 16 y^ -\- gx^ = 144. What 
is the distance of its foci from the center, and the distance of 
each focus from the vertices ? 

5. The distance from the focus to the end of the conjugate 
axis of an ellipse is 5, and its eccentricity is f . What is the 
equation of the ellipse and the distance of the focus from the 
vertices ? 

6. The semi-conjugate axis of an ellipse is 8, and the dis- 
tance from its focus to its center is 6. What is the equation 
of the ellipse ? Where does the ellipse cut the circle whose 
radius is 9? 



PROPOSITION VI 

131. If r' be the longer and r be the shorter focal radius of any 
point on an ellipse^ theri 

r' =^ a-\- ex, 
and r=:i a — ex, 

in which x is the abscissa of the point, e the eccent7'icity , and a 
the semi-transverse axis of the ellipse. 

Let r' = F'P and r^ FP, 
^^CK *' ^ = PK, 
<f^CF '* ^ = CA, 
and e =the eccentricity. 




87 



Fig. 40 



[i] Then r'' = PK + F'K, 

[2] but F'K = CK+CF'=:ji: + ^. 

[3] Hence r" = f + {x + c)\ 



bv Geom. 26. 



[4] 

[5] But 

[6] Hence 



by Geom. 26. 



FK = CK — CF = X — c 

r' = y + (jc—cy\ 

By subtracting the members of [6] from the corresponding 
members of [3], we get 

[7] r" — r'= (x + cy—{x — cy=^€X. 

[8] Hence {^ -\- r) (/ — r)=4cx. 

r' + r = 2a. 
2a {r^ — r) = 4 ex, 
c 
a 

czzzae. 
2a (r' — r) = ^.aex, 
r' — r= 2ex. 
r'-\-r^=2a. by § 109, 



[9] But 

[10] Hence 

[11] But 

[12] Hence 

[13] Hence 

[14] and 

C15] 



by § 109. 
by § 128. 



By adding [14] and [15] 

[16] 2r' = 2^ -|- 2^x, 

[17] and /=<2 + <?jr. 



88 



ANALYTIC GEOMETRY 



By subtracting [14] from [15] 

[18] 2r=2« — 2ex. 

[19] r ^=. a — ex. 

132. The Secant. — A secaiit is a straight Hne cutting a 
curve in two points. 

If one of the two cutting points remains fixed and we 
make the other move along the curve till the moving point 
coincides with the fixed point, the secant will revolve about 
the fixed point as a pivot. 

133. The Tangent. — When the two points in which the 
secant cuts the curve coincide, the secant is called a tange7it. 

PROPOSITION VII 

134. The equation of the tangent to an ellipse is 

y—y = — ^r (^ —^) , 

in which x^ and y' are the coordinates oj the point of tang ency 
and a and b are the semi-axes. 




Fig. 41 



I^et P'T be a tangent to the ellipse at the point P'. 
Let.r' = CK andy=P'K, 
I.et« =CA " ^=CB. 



THE ELLIPSE 89 

We are to prove that 

is the equation of the tangent to the ellipse. 

Let P'M be a secant cutting the ellipse at the two points P' 
and P". 

Let jr"=CH andy" = P"H. 

Since the secant is a straight line passing through the two 
points P' and P", its equation must be of the form 

[i] y__^:=Z'_^|(^'_^). by §58. 

Since P' is on the ellipse, its coordinates x^ andj^' must sat- 
isfy the equation of the ellipse. by § 40. 

Substituting x^ and ^ for the x and y in that equation, we 
get 

[2] d^y^^ -{- b'x^''^ ■==■ a~b\ by § III. 

Similarly, since P" is on the ellipse, its coordinates ;tr" and 
jj^" must satisfy the equation of the ellipse. 

Substituting x^^ and j/" for the x andj/ in that equation, we 
get 

[3] ay" + b'x'" = arb\ 

Subtracting the members of [2] from the corresponding 
members of [3], we get 

[4] a\y'^—y'') -\-b\x"' — x") =0. 

[5] b\x''''-x") =-a\y"-y"). 

[6] b'^ {x' + y ) (:r" -x')=-d {y' +/' ) ( /' -/ ) . 



[7] 



x'^—x'~ a"{y' + y")' 



^ . . . y" —y' . 

Substituting this value of-^7^ — =^, into [i], we get 

[8] •^-^ = - ^^(y+y) (^-^)' 

Now let the point P" move along the ellipse towards P'. 



90 ANALYTIC GEOMETRY 

Then the secant P'M will revolve about the point P' and will 
continually approach the tangent P'T, and when P" reaches 
P' it wnll coincide with the tangent. But when P" reaches 
P' we will have 

[9] ;f" = ji;' andjj/" ^ jl/', 

and the fraction in [8] will become 



[10] 



•2 t — 2 I' 
a 2y ay 



Substituting this value of the fraction into [8], we get 

b'' x^ 
[11] -^'~"-^=~^^-^'~-^)- 

Now, the X and y of [8] stand for the coordinates of every 
point on the secant in every position which it takes as it re- 
volves about P'. Hence they stand for the coordinates of 
every point on it when it coincides with the tangent. 

But when the secant coincides with the tangent, [8] takes 
the form of [11]. Hence the x and y of [11] stand for the 
coordinates of every point on the tangent. 

Therefore [11] is the equation of the tangent. by § 39. 

Q.E.D. 

b'' x^ 

135. Corollary i . — The fraction -^ — ^ is the slope of the tan- 
gent. 

y"—y' 
For the fraction -^ „ , is the slope of P'M, by § 59, and 

<^' x' v" — y' 
5 — i is the form that =^, , takes when the secant com- 

a^ y' x" — x' 

cides with the tangent. 

136. Corollary 2 . — The equation of the tangent to a circle is 

x' 

y'—y=z — y{x' — x). 

For when the ellipse takes the form which we call the circle 
b=^a and [11] becomes 

x' 

y'—y=—-y{x'—x). 



THE ELLIPSE 91 

137. The Subtangent. — The subtangent is the distance 
measured along the X axis, from the ordinate of the point of 
tangency to the tangent. 

138. Corollary j. — The length of the subtangent of an ellipse is 
a — X 



x' 



For, since the point T is on the tangent P'T, its coordinates 
X = CT and j^/ = o, must satisfy the equation of the tangent. 

by §40. 
Substituting CT and o for the x and y of that equation, 

we get 

b'' x^ 

[i] y=-V^,(y-cT), 

ay 
[2] ay = — b'x'' + b'x'CT, 

[3] «y^ + ^V^ = <^VCT. 

Since the point P' is on the ellipse, its coordinates x' and j/' 
must satisfy the equation of the ellipse. by § 40. 

Substituting x^ and j/' for the x and y of that equation, we 
get 

[4] ay + b'x'' = a'b\ 

[5] Hence b'x^CT = a'b\ by [3] . 



[6] CT 



a' 



x^' 



[7] and KT =CT — CK = -^ — x^ 



a^ . a^ — y^ 



X' x' 



139. Corollary 4.. — When the ellipse takes the form which we 
call the circle^ 



the subtangent = ^^ . 

For then 
[i] a—r, 

a' — x'' r' — x" y" -^ , ^ 

[2] and ^, = ^, =~x^\ by § 138. 

Draw a figure showing all the lines referred to in this cor- 
ollary. 



92 . ANALYTIC GEOMETRY 

140. Co7vllary 5. — If different ellipses have the same trans- 
verse axis, and ordinates be drawn to each from the same point on 
the transverse axis and tangejits be drawn at the extremities of 
these ordinates, then all the subtangents will be equal to each 
other. 

For in the case of each ellipse 

the siibtangent = ^ — . by § 138. 

Draw a figure showing all the lines referred to in this 
corollary. 

141. Corollary 6. — To draw a tangent to an ellipse from any 
given point on it. 

Circumscribe a circle about the ellipse, as in § 116. Draw 
an ordinate, cutting both ellipse and circle. Draw a tangent 
to the circle at the point where the ordinate cuts it. Join the 
point where this tangent cuts the transverse axis produced to 
the point where the ordinate cuts the ellipse. This last line 
drawn will be the tangent required. 

Draw a figure and give the proof. 

EXAMPLES 

Required the equation of the tangent to and find the sub- 
tangent of each of the following ellipses : 

1. 2x'^ -\r 4-y^ ^= 38, when (1,3) is the point of tangency. 

2. jc- + 4r-= 20, " (2, 2) " " " " 

3- -^+^ = I, " (^.n) " - - - 
m n 

142. The Normal. — The normal to a curve is a straight 
line perpendicular to the tangent at the point of tangency. 

PROPOSITION VIII 

143. The equatio7i of the normal to an ellipse is 



THE ELLIPSE 



93 



in which x' and y' are the coordinates of the point of ta^igency and 
a and b the semi-axes. 




Fig. 42 

Let PN be the normal and PT the tangent to the ellipse at 
the point P. Draw the ordinate PK. 

Let x' = CK and 7' ~ PK. 
Let .y' = the slope of the tangent PT. 

The normal is a. straight line passing through a fixed point, 
namely, the point of tangency. by § 142. 

• Hence its equation must be of the form 

[i] y~-y::^s{x'~x), by §57. 

in which x' and y' are the coordinates of the fixed point, 
here the point of tangency, s is the slope of the line, that 
is, the slope of the normal, and x and y the coordinates of 
any point on the line, here the coordinates of any point on 
the normal PN. 

But the normal is perpendicular to the tangent. by § 142. 

by § 62. 

by § 135- 



= o. 



^2_ 


Hence 


I -j- ^^' = 0, 


[3] 


but 


d'x' 
^ ~ a'y'' 


[4] 


Hence 


, / b' x'\ 


'"' A a^yJ- 


[5] 


Hence 


a'y 



94 ANAL YTIC GEO ME TR Y 

Substituting this value of ^ into [i], we get 

[6] y - y ^"11^ ^^' - x^ . 

Now, in [6], X andj/ are the coordinates of any point on 
the normal PN. Hence [6] is the equation of the normal. 

by § 39. 
Q. E. D. 

2 f 

144. Corollary i. — The fraction j^-^ is the slope of the normal. 

145. The Subnormal. — The subnormal is the distance 
measured along the transverse axis from the ordinate of the 
point of tangency to the normal. 

h' 

146. Corollary 2. — The length of the subnormal is —^x'. 

For in Fig. 42 the point N is on the normal, hence its co- 
ordinates jm= o and X = CN must satisfy the equation of the 
normal. 

Substituting these values of x and y into the equation of 
the normal, we get 

[I] y-o=^,(x'-CN). by §143. 

2 12 

[2] Hence CN = ^ ~ x'. 

-■ a 

[3] But NK^CK — CN==y — "^ ~^ x', 

[4] Hence NK =- ^x' . 

146(2 . Corollary ^. — The intercept of the normal on the X axis 
is equal to e'^x'. 



For 

a'—b' 



a 
Hence [2] becomes 

CN = e'x'. 



= e\ by §128, [4]. 



THE ELLIPSE 95 

147. Corollary 4.. — The equation of the normal to the circle is 






X 



For when the ellipse takes the form which we call the cir- 
cle, b^^ a and the equation of the normal becomes 

V 
EXAMPLES 

1 . A straight line touches the ellipse whose semi-axes are 
4 and 3 at the point whose coordinates are — 3 and 1.9. 
What is its slope ? 

2. A line whose inclination is 45"" touches 25^ -\- i6j»r^ = 
400. What are the coordinates of the point of tangency ? 

Ans. X = — 3.9. 
y = 2.s. 

3. Find the equation of the tangent drawn to i6y^ -\- gx^=i 
144 at the extremity of the parameter. 

Ans. _y = — 0.67^1; 4" 4.04. 

4. Find the equation of the tangent to the ellipse a'y^ -\- 
b^x' = a'^b'^ at the extremity of the parameter. 

5. A straight line touches 25 jk^ + i6jc^ = 400 at the point 
whose coordinates are — 3, 3.2. What are the angles which 
it makes with the axes ? What is the area of the triangle 
betw^een it and the axes ? 

6. A tangent to a~y^ + U^x"^ = a^b^ makes equal angles with 
the axes. What are the coordinates of the point of tangency ? 

Ans. X ^ 







a' 




Va 


y 




' + b' 
b' 


y — 


'Va 
the 


• 


.00 at 


'+b' 

point 



7. A normal is drawn to 2^y'^ -\- i6;t:^ 
whose coordinates are — 3 and 3.2. What are the angles 
which it makes with the axes ? What are its intercepts ? 
Find the distances measured along the normal from the point 
of tangency to each of the axes. 



96 



ANALYTIC GEOMETRY 



Find the equation of the normal and the value of the sub- 
normal in each of the following ellipses : 

8. 3jK^ + 4jr^ = 39, when (3, i) is the point of tangency. 

9. jc' + 4y = 2o, . " (2,2) " " " " 

10. ay+^v=a"^' " («, o) " " " " 



PROPOSITION IX 



148. In an ellipse the normal bisects the inteHor^ and the tan- 
gent the exterior a7igle between the focal radii of the point of 
tangency. 




Fig"- 43 

Let PN be the normal and PM the tangent to the ellipse at 
the point P. 

Let FP and F'P be the focal radii of the point P. 
We are to prove that the normal PN bisects the angle FPF' 
and that the tangent PM bisects the angle FPQ. 



[I] 
[2] 
[3] 
[4] 

[5] 
[6] 



CN = ^V. 



§ i46<2. 



NF=CF— CN = r — ^V = «^ — ^V. by § 128, [2]. 

Hence NF --^ e{a — ex^) . 

Similarly NF' = CF' + CN = ^(^ + ^jr'). 

NF _ a — ex' 

NF ■" a + ex'' 

NF r 



Hence 



and 



NF' 



by § 131. 



THE ELLIPSE 97 

[7] or NF : NF' : : r : r\ 

Take QP = PF = r and draw QF. 
Substituting this value of r into [7], we get 

[8] NF : NF' : : QP : r\ 

Hence is QF || to PN. by Geom. 24. 

[9] Hence ^ F'PN = z^! PQF, by Geom. 8. 

[10] and ZL NPF == Z PFQ. by Geom. 7. 

[11] But ^Pi^Q = ^PQF, by Geom. 16. 

[12] hence Z. F'PN = z^ NPF. 

Q. K.D. 

Again 

[13] Z^F'PR+ZlF'PN = z^FPN + ^FPM, by§i42. 

[14] and ^F'PN = z::FPN, by [12]. 

[15] hence zl F'PR = zl FPM. 

[16] But Z. F'PR = Z. QPM, by Geom. 4. 

[17] hence z:! QPM = zlFPM. 

Q. E. D. 

149. Corollary i . — To draw a tangent to an ellipse at a given 
point on it. 

Draw focal radii to the given point. Produce either of 
these focal radii and bisect the exterior angle between them. 
The bisector is the tangent. Draw a figure and give the 
proof. 

EXAMPI^ES 

1. Focal radii are drawn to the point 3, 3.4 on the ellipse 
Z^y^ + i6jt:^ = 576. Find the length of the focal radii and 
the angle between them. 

2. Focal radii are drawn to the point 3, 3.2, on the ellipse 
25jj/' + 16^^ = 400. Find the angle which they make with a 
tangent to the ellipse at the same point and the length of the 
perpendicular drawn from the focus to the tangent. 

150. A Chord. — A chord of an ellipse is a straight line 
terminated both ways by the ellipse. 



98 



ANALYTIC GEOMETRY 




Fig 44 

Let RS, R'S', and R"S" be parallel chords, and P, P', and 
P'' their middle points. 

lyet MT and M'T' be two tangents parallel to these chords. 

Conceive the number of parallel chords between MT and 
M'T' to be increased continually until they touch each other 
side by side. Then their middle points will also touch each 
other and form a continuous line. Such a system is called a 
complete system of parallel chords. 

151, The Bisector of a Complete System of Parallel 
Chords. — The bisector of a complete system of parallel chords is 
the line which contains all the middle points of those chords. 

152, The Diameter. — The diameter of an ellipse is that 
part of the bisector of a complete system of parallel chords 
which is terminated both ways by the ellipse. 

PROPOSITION X 



153. The equation of the diam,eter of an ellipse is 



y 



( ^ cot cp)x, 



a 



in which cp is the inclinatio?i of the system of chords bisected by 
the diameter, and a and b are the semi-axes. 

Let RS represent any one of a complete system of parallel 
chords, cp its inclination, and P its middle point. 




99 



]l' 




[< 


] But 


[3: 


] Hence 


[4: 


1 and 


[5: 




[6; 


] Hence 


[-: 


] and 


[«: 




[9: 


Hence 


[lo; 


] and 



by Geom. 8. 



by Trig. 23, 
by Trig. 2. 



Let jr = CK and j = PK, 
y = CH " y = RH, 
and r= PR = PS. 

Draw PO 1| XX'. 

z:RPO = ^PMK. 

Z^PMK= 180— (?^. 

z:: RPO = iSo—cp^ 

cos RPO = — cos ^. 

PO = r cos RPO. 

Hence by [4] PO= — r cos (p, 

XI = CK — PO = x-\- r cos (p. 
RO= rsin RPO. by Trig. i. 

RO := r sin cp, by Trig. 22. 

J'' = VK + RO = y + r sin <7?. 

Now the point R is on the ellipse, and hence its coordinates 
x' 2iVi& y' must satisfy the equation of the ellipse. by § 40. 

Substituting x^ and jk' for the x and_y in the equation of the 
ellipse we get 

[it] a^f''^b'x''^a'b\ by §111. 

Substituting the values of x' andy found in [7] and [10] 
into [11], we get 

[12] a'^iy -\- rsin cpY -\-b^\x +r cos cp)^ = a'^b'^. 



loo ANALYTIC GEOMETRY 

Squaring the binomials and factoring with respect to r^ and 
r, we get 

[13] {a^ sin^ ^~\- b' cos^ q))r'^ -^ 2 {ay sin (p-\- <^'Jt: cos (p)r=: 

a'd' — ay'—d'x'. 

Now, since P is the middle point of RS, the two values of 
r in [13] must be equal to each other, and hence by the theory 
of quadratic equations 

[14] 2 (ay sin cp + ^'-^ cos (p) =^ o. 

[15] ay sin cp =1 — d^x cos cp. 



[16] y=\ r- — j-^> 

^cot ^1^. by Trig, 9 and 6. 

Now the X and jf of [17] stand for the coordinates of the 
point P. But since RS represents any one of the complete 
system of parallel chords, P is any point on their bisector, 
and since the diameter lyN is a part of the bisector, P is any 
point on that diameter. 

Hence the x andjF of [17] stand for the coordinates of any 
point on the diameter which bisects the system of chords rep- 
resented by RS, and therefore [17] is the equation of that 
diameter. by § 39. 

Q. E. D. 

154, Corollary I . — The diameter of an ellipse is a straight line 
passing through the center. 

lyCt the chord RS move across the ellipse, always remain- 
ing parallel to itself. Then its middle point P wall trace out 
the diameter which bisects the system of chords represented 
by RS. 

Now as P moves along the diameter, a, b, and cp always re- 
tain the same value and therefore are constants. by §5. 

Hence the expression ^ cot <z^ is a constant. Now if we 

a~ 

represent this constant by s, [17], which is the equation of 

the diameter, may be written 



THE ELLIPSE 



lOI 



[i8] _)' = sx. 

But [i8] is the equation of a straight line passing through 
the origin. by § 55. 

Hence, since the origin is at the center of the ellipse, the 
diameter is a straight line passing through the center. 

155, Corollary 2. — If 6 is the inclination of a7iy diameter, 
and (p the inclination of its system of bisected chords, then 

b' 



tan 6 tan ^ = — 



a 



Let be the inclination of any diameter. Then tan 6 is 
its slope. by § 52. 

Since by Corollary i, the diameter is a straight line, the 
coefficient of x in its equation must be the slope of the diam- 
eter. 



[I] 
[2] 
[3] 



Hence 



But 



tan a = ^ cot cp. 



a 



cot q) = 



tan cp 



Hence tan tan ^ = — 



b' 



a' 



by § 53. 
by § 153- 

by Trig 9. 



Q. E.D- 



PROPOSITION XI 

^5^' ^J ^^y diameter bisect a system, of chords which are 
parallel to a second diameter, then that second diameter will bisect 
a system parallel to the first diameter. 

Y 




I02 



ANALYTIC GEOMETRY 



Let the diameter MN bisect a system of parallel chords 
represented by RS, and let the diameter OP bisect the sys- 
tem represented by TU, and let RS be parallel to OP. 

We are to prove that TU 1| MN. 

Let Q be the inclination of MN. 

( > ^yj ( ( ( ( 4 ( 

i ( Cl\ ( ( ( ( ( t 



9 



f i i ( ( 



[i] Then 



[2] and 



tan tan ^ = — 
tan 6' tan ^' = — 



" RS. 




" OP. 




" TU. 






by§ 155. 


b' 




a'' 




>' tan cp\ 






by hypoth. 




by Geom. 8. 



[3] Hence tan d tan cp^ tan Q' tan cp^ . 

But RS is parallel to OP. 

[4] Hence ^' = <p, 

[5] and tan ^' = tan cp. 

Substituting this value of tan q) into [3], we get 
[6] tan Q tan & — tan d' tancp'. 

[7] Hence tan 6 = tan cp', 

[8] and = cp'. 

Hence TU |1 MN, by Geom. 9. 

and the system of chords parallel to TU will be parallel to 
MN. by Geom. 10. 

Therefore OP bisects a system of chords which are parallel 

to MN. 

Q. E.D. 

157. Conjugate Diameters. — Two diameters are said to be 
conJ2igate to each other when each bisects a system of chords 
which are parallel to the other. 

158, Corollary i . — If be the inclinatio7i of any diatneter, and 
6' the inclination of its conjugate^ then 



tan 6 ta7i 0' = ^. 

a~ 



THE ELLIPSE 



lo- 



[i] For 
[2] But 
[3] Hence 



tan Q tan ^ =^ — 
tan qj = tan Q\ 



a' 



tan Q tan Q' — ^. 

a' 



by § 155- 
by § 157. 



PROPOSITION XII 



159. The tangent to an ellipse at the extremity of any diam- 
eter is parallel to the C07ijugate of that diaTneter. 




Let OP be any diameter and MN its conjugate. 
Let PT be the tangent at the extremity of OP. 
We are to prove that PT |! MN. 

Let RS be any one of the system of chords bisected by OP. 

Let Q be the inclination of OP, 



d 
9 



/ ( ( ( ( 



( ( ( ( 



" MN, 
" RS, 
" PT. 



and ^' '' " 

Let y = CK andy = PK. 

RS will be parallel to MN. 
[i] Hence cp = 0\ 

[2] and tan ^ = tan 0'. 



by § 157' 
by Geom. 8. 



I04 ANALYTIC GEOMETRY 

Since the point P is on the diameter OP, its coordinates x^ 
and y must satisfy the equation of that diameter. by § 40. 
The equation of the diameter OP is 

[3] JV= (— ^cot ^)jf. by § 153. 

Substituting x^ and j^' for the x and y of this equation, we 
get 

b' 



[4] 




y=(- 


, cot cp^x\ 




[5] 


hence 


cot ip 


ay 

~ b'x'' 




[6] 


But 


cot q) 


I 


by Trig. 9. 


tan (p' 


[7l 


Hence 


I 


" 1 
ay 

■~ b'x^ ' 




tan (p 




[8] 


and 


tan (p 


_ b'x^ 

— •' (• 
ay 




[9] 


Hence 


tan 0' 


b'x^ 


by [2]. 


— 1 • 
ay 


[10] 


But 


tan (p* 


b'x' 


by § 135. 


~ ay ' 


[II] 


Hence 


tan^' 


= tan (p\ 


by [9] and [10]. 


[.2] 


and 


(9' 


= cp'. 








Therefore PT || MN 


by Geom. 9. 










Q. E. D. 



160. Corollary i . — The two tangents at the extremities of a 
diameter are parallel to each other. 

161. Corollary 2. — The four tangents at the extremities of two 
conjugate diameters form a parallelogram circumscribed about the 
ellipse. 

PROPOSITION XIII 

162. Given the coordinates of one extremity of a?ty diameter of 
an ellipse to find the coordinates of the extremities of its conju- 
gate. 



THE ELLIPSE 



105 



X' 







Y 


t\ 




R 


-^^^ 






1 






.> 


\t 


\ " /^ 


Nv L 


; 


\ 


qV^^__ 




/ 
1 








V 







Fig 48 

Let PQ be any diameter and RS its conjugate. 

Let OK and PK be given. 

We are to find OH and RH, and OL and LS. 

Let y = OK and/ = PK. 
*' y ^ OH andy = RH. 

Draw PT tangent to the ellipse at the point P. 

Since RS is a straight line passing through the origin, its 
equation must be of the form 



[I] 




y = SX. 


by § 55. 






But RS 11 PT. 


by § 159. 


[2] 


Hence 


z:rox = z:ptx, 


byGeom.8. 


[3] 


and 


tan ROX = tan PTX. 




[4] 


But 


tan ROX = s, 


by [i] and § 53. 


[5] 


and 


b^x' 

tan PTX = Y-n 

ay 


by § 135. 


[6] 


hence 


b'x' 




ay 





If we substitute this value of s into [i], the equation of RS 
becomes 

[7] y = ^x- 



a'y' 



io6 ANALYTIC GEOMETRY 

Now since the point R is on the diameter RS, its coordi- 
nates x" and J/" must satisfy the equation of RS. by § 40. 
Hence substituting jf" andjF" for the x andj/ of [7], we get 



[8] y = - 



a'y 



The equation of the ellipse is 

[9] a-y'^ ■\- b^x'^ ■=! a^b- , by § iii. 

and since the point R is on the ellipse, its coordinates :r" and 
j/' must satisfy the equation of the ellipse. by § 40. 

Hence substituting jf" and y" for the x andy of [9], we get 

[10] ay" + b'x"'=^a'b\ 

Now since in both [8] and [10], x" stands for OH and y 
stands for RH, these equations are simultaneous, and there- 
fore can be solved by algebra. 

Squaring both members of [8], we get 

b'x''^ 

Substituting this value of y' , into [10], we get 

[12] a' ^-^,x"' + b'x'" = a'b\ 

-■ ay' 

[13] l^x'- + X'- ^ a\ 

[14] b'x"x"' + ay'x'" = ay\ 

[15] {ay + b'x")x"' = ay\ 

Now since the point P is on the ellipse, its coordinates x' 
and y must satisfy the equation of the ellipse. 

Substituting x' andy' for the x andjK of [9], we get 

[16] ay' + b'x" = a'b-\ 

Substituting the value of the left hand member of this 
equation into [15], we get 

[17] a'b'x"'^ay\ 



[r8] Hence x" = =ir -^y. 



THE ELLIPSE 107 

That is, OH = — -^ y' and OL = -^ y'. 
b b -^ 

Substituting -\ — j-y for ^" in [8], we get 

[19] y — — -— y — x' — LS. 

ay b a 

Substituting y-j/' for ;f" in [8], we get 

r -1 ,/ ^^-^' a , b , ^^^ 

[20] jf/" = —^, -j-y' = ^x'=: RH, 

ajK b '^ a 

163. Corollary. — All diameters are bisected by the center. 

[i] For OL==OH, by [18]. 

[2] and LS = RH. by [19] and [20]. 

[3] Hence OS = OR. by Geom. 15. 

EXAMPLES 

1. A straight line touches 36^ + i6.r^ ^ 576 at the point 
whose coordinates are 3 and 3.4. What is the equation of 
the diameter conjugate to the one which passes through the 
point 3, 3-4? 

2. The inclination of a diameter of 25^+16^^ = 400 is 
45''. What is the equation of the tangent at the vertex ? 

Ans. y——\tx^ 32. 

3. A straight line touches iSy^ -\- gx^ ^= i4./\. at 2, 2.6. 
Where does the diameter which is parallel to this line cut the 
curve? Ans. At — 3.1, 1.5, 

and at 3.1, — 1.5. 

4. A diameter cuts 2sy' + i6x^ = 400 at — 3, 3.2. What 
is the equation of the tangent parallel to that diameter ? 

PROPOSITION XIV 

164. The sum of the squares of any pair of co7ijugate diain- 
eters is equal to the sum of the squares of the axes. 



io8 



ANALYTIC GEOMETRY 
Y 




I/Ct PQ and RS be any pair of conjugate diameters, and 
AA' and BB' the axes. 
We are to prove that 

2 2 ^ 2 2 

PQ + RS = AA'+BB'. 

■ Lety ^CKandy ^PK. 
;»;" = CH " y' = RH. 
a ^AC " ^=BC. 



a' 



CP 



^'=CR. 



[0 




a"- 


= x" +y^ 


by Geom. 26. 


[2] 




b" = 


= x'" + y"\ 


by Geom. 26. 


[3] 


But 


112 


4-^^ 


by § 162, [18], 


[4] 


and 


y" = 




by S 162, [20]. 


[5] 


Hence 


b" = 




by [2]. 



Adding [i] and [5], we get 
[6] 



a" + 6'^ = ^"' + ^ ^" + -^y"+ y\ 



[7] and a" + ^ = {a'+ f) "^ + {a' + b') ^, 



THE ELLIPSE 109 

[8] „.^ + ^-=(«« + ^^)g+Z!). 

Now since the point P is on the ellipse, its coordinates x^ 
andjj/' must satisfy the equation of the ellipse. by § 40. 

Substituting jr' and jj/' for the x andj^in that equation, we get 

[9] a'y' + b'x^' = a'b\ by § 1 1 1 . 

[10] hence ~j^~\~~^^^^- 

Substituting the value of the left hand member of this equa- 
tion into [8], we get 

[11] a^'' -\- y' ^ a" -\- b\ 

[12] Hence 4^'^ + 4<^'"= 4^^+ 4<^^- 

[13J But a' — iPQ, b' — iRS, a — ^AA' and b — ^BB'. 

by § 163. 

2 2 2 2 

[14] Hence PQ + RS = AA' + BB'. by Geom. 32. 

Q. E. D. 

PROPOSITION XV 

165 . The parallelogram formed by tangents to an ellipse at the 
extremities of any pair of conjugate diameters is equivalent to the 
rectangle whose sides are equal to the axes of the ellipse. 

Let PQ and RS be any pair of conjugate diameters, and let 
L,MN O be the parallelogram formed by tangents at their ex- 
tremities. 

Let DBFG be the rectangle whose sides are equal to the 
axes AA' and BB'. 

We are to prove that 

LMNO = DEFG. 

Let« ^CA and^=CB, 

a' =CP " <^'=CR, 

jr' =CH " y =PH, 

x" = CK " y'=RK, 

o=^pcH** 99 =z:rch. 



no 



ANALYTIC GEOMETRY 




[i] LMNO = LO X NO sin I.ON. by Trig. 15. 

LOURS, by § 159- 

and NO 11 PQ. by § 159. 

Zl lyON = ^ RCP, by Geom. 1 1 . 

sin I,ON = sin RCP = sin ^cp — d). 

\,0— RS. by Geom. 17. 

■ RS= 2^'. by § 163. 
LO = 2b\ 
NO = 2a\ 



[2] Hence 

[3] and 
[4] 

[5] 

[6] Hence 

[7] Similarly 



Substituting these values of sin LON, lyO and NO into [i], 
we get 

[8] LMNO = \a'b' sin {cp — d). 

[9] sin ((^ — ^) = sin ^ cos — cos ^ sin ^. by Trig. 13. 
sin cp = sin RCK. by Trig. 22. 

_ RK _ y 



[10] 

[11] But 

[12] Hence 
[13] 



sin RCK 



RC 



by Trig. i. 



rs CH x^ 

COS0 = ^p=^ 



by Trig. 2, 







THE ELLIPSE 


Ill 


[14] 




cos ^= — cos RCK. 


by Trig. 23. 


[15] 




COS RCK = CR = ^. . 




[i6] 


hence 


cos q)— ^, . 




[17] 




. ^ PH y 





Substituting these values of sin cp, cos Q, cos <?? and sin Q 
into [9] , we get 

[18] sm(^-0) = f ^ + ^^=— ^^. 

[19] But •^" = ^y, by §162, [18]. 



[20] and jf" = — •^'. by § 162, [20]. 

r -1 XT w ^x ~^^ ^"^ ^v^ + «y^ 

[21] Hence sin (^ — 0) — ~ — 



a'd' ~ a'd'ad ' 

Since the point P is on the ellipse, its coordinates jr' and_>'' 
must satisfy the equation of the ellipse. by § 40. 

Substituting x' and_y' for the x and y in that equation, we 
get 

[22] ay' + d'x" = a-d\ 

Substituting the value of the left hand member of this 
equation into [21], we get 

Substituting this value of sin {cp — 0) into [8], we get 

[24] LMNO = 4.a'd^ — ^ = ^ad. 

[25] But 4.ad= 2a2d — AA' X 'BB\ 

[26] and AA'=GF. by Hypoth. 

[27] Hence 4a^=:GF X BB' = DEFG. by Geom. 28. 



112 ANALYTIC GEOMETRY 

[28] Therefore I.MNO = DKFG. 



by [24]. 
Q. K. D. 



i66. Corollary. — The sides of the rectangle circumscribed abotd 
an ellipse are equal to the axes. 



PROPOSITION XVI 
167. The area of an ellipse is n times the product of the semi- 



axes. 




Let i^'^the area of the ellipse. 
" «^CA and /^ = CB. 

We are to prove that 

E = nab. 

Circumscribe a circle about the ellipse. 

Draw ordinates to the circle, cutting the ellipse in the 
points P, Q, and R. 

Join the points P and Q, Q and R, R and B, S and T, T 
and W, W and V. 



[I] 



ML 



PQML = (PI. + QM) -^, by Geom. 61. 



[2] and STML=(SL + TM) 



2 
ML 



by Geom. 61. 



THE ELLIPSE 113 

[3] Hence ^^^ P^li Q^ 

L3J ^ence sTML SL + TM' 

[4] PL : QM : : SL : TM. by § 119. 

[5] Hence PL : SL : : QM : TM, by alternation. 

[6] and PL + QM : SL + TM : : PL : SL. by Geom. 22. 

[7] But PL:SL::^:a. by §118. 

[8] Hence PL + QM : SL + TM : : ^ : «, 
r -1 PL + QM _ /^ 



LVJ 


KJL 


SL + TM a' 


Substituting the 


right hand member of this equation into 


[3]> we 


get 




[10] 




PQML b 
STML"~ a ' 


[11] 


Similarly 


QRNM_ d 
TWNM ~ a ' 


]l2J 


and 


RBCN b 
WVCN ~ a ' 


[13] 


Hence 


PQML QRNM RBCN 
STML ~TWNM~ WVCN' 



[14] or PQML+ QRNM + RBCN: STML + TWNM + 
WVCN :: PQML : STML. by Geom. 22. 

Let C = the area of the circumscribing circle. 

2Tc =^ the sum of the trapezoids in the quarter of the 
circle and 

2Tq= the sum of the trapezoids in the quarter of the 
ellipse. 

Then [14] will become 

[15] :ST, : :^T, : : PQML : STML, 

r Ai ^^e PQML b - K r n 

[16] or ^^=STML = V ^^^'''^' 

[17] Hence a2T^:= b^T^. 



114 



ANALYTIC GEOMETRY 



Now let the number of trapezoids be increased continually. 
Then the two members of [17] will be variables. by § 5, 

[18] Hence limit ^-^Z'e = limit <^ -2" 71-. by Geom. 20. 

But while the number of trapezoids increases, the a and b 
of [17] always retain the same values. They are, therefore, 



constants. 

[19] Hence a limit ^T^^^ b limit ^T^. 

E 



by § 4, 



JIO 




lyimit -^/"e = — • 
4 


by Geom. 19. 


[21] 




lyimit ^T^ =z — , 

4 


by Geom. 19. 


[22] 


Hence 


a — := b — , 
4 4 


by [19; . 


[23; 


and 


aE ^= bC. 




[24] 


But 


C = 7^<2^ 


by Geom. 30. 


\2<\ 


Therefore E = nab. 





Q. B. D. 



PROPOSITION XVII 



168. If the inclinations of two diameters be supplementary 
angles, the diameters must be equal to each other. 




Fig. 52 



Let be the inclination of the diameter RS, and ^' be the 
inclination of the diameter PQ. 

Let 0' = 180° — e. 





THE ELLIPSE 


"J 


We are to prove that 






RS = PQ. 




[I] 


d' = 180° — e. 


by Hypoth 


[2] Also 


6' - iSo° ^PCA'. 


by Geom. 3 


[3] Hence 180° - 


e - 180° /pcA', 




[4] and 


e = /pcA'. 





Now about BB' as an axis revolve BAB" till it comes into 
the plane of BA'B'. 

Then since 

z:bcai=z:bca', 

CA will take the direction of CA'. 
And since 

^-z:pca', 

CR will take the direction of CP. 

Now ARB and A'PB are symmetricar with respect to BB'. 

by § 126. 

Hence R will fall upon P, and therefore CR will coincide 
with CP. 

Therefore RS = PQ. by § 163. 

Q. E. D. 

PROPOSITION XVIII 

i6g. The two conjugate diameters whose inclinations are sup- 
plementary angles are^ when produced^ the diagonals of the rectan- 
gle formed on the axes. 

Let RS and PQ be two conjugate diameters whose inclina- 
tions Q and ^' are supplementary angles. 

Let LMNO be the rectangle on the axes. 

We are to prove that RS and PQ produced are the diagonals 
of the rectangle LMNO. 

Since RS and PQ are conjugates 
[i] tan ^ tan ^' = ^. by § 158. 



Ii6 



ANALYTIC GEOMETRY 




Fig. 53 

But since d and ^' are supplementary 

[2] —tan (9 — tan ^'. by Trig. 24. 

Substituting this value of tan ^' into [i], we get 

tan^ e = 



[3] 

[4] 
[5] 
[6] 
[7] 



a 



2 



Hence tan^= d= 



<2 



or 



But 



or 



h h 

tan RCA = — , and tan RCA' ~ 



a 



a 



tan LCA = 
tan LCA = 



LA 
CA' 

b 
a 



by Trig. 3. 



Hence by [5] and [7] we get 

[8] tan RCA := tan LCA. 

[9] Hence ZRCA^^LCA. 

Therefore the diameter RS coincides with the diagonal LN. 

Similarly it may be shown that the diameter PQ coincides 
with the diagonal OM, 

Q. E. D. 

170. Corollary 1 . — The conjugate diameters whose inclinations 
are siipplemejitary angles are equal to each other. by § 168. 

171. Corollary 2. — There are only two conjugate diameters 
whose inclinations are supplementary angles. 



THE DIRECTRIX 



117 



For there are only two diagonals of the rectangle formed on 
the axes. But the conjugate diameters whose inclinations 
are supplementary coincide with these diagonals. by § 169. 

Hence there can only be two such conjugate diameters. 

172. Equi-Conjugate Diameters.— The two conjugate 
diameters which, when produced, are the diagonals of the 
rectangle formed on the axes are called the equi-conjugate 
diameters of the ellipse. 

The Directrix 

173. The Directrix. — The^2y^^/r2.r of an ellipse is astraight 
line drawn perpendicular to the X axis on the opposite side 
of the focus from the vertex, and at such a distance from the 
vertex that the distance from the focus to the vertex divided 
by the distance from the vertex to the perpendicular is equal 
to the eccentricity of the ellipse. 




Fig. 54 



In Fig. 54 if 



FA 
AE 



e, then DD' is the directrix. 



174. Focal Radius. — The distance from au}^ point on the 
ellipse to the focus is called \h.^ focal radius of that point. . 

175. The Directral Distance. — The distance from any point 
on the ellipse to the directrix is called the directral distance 
of that point. 

PD is the directral distance of the point P. 



ii8 



ANALYTIC GEOMETRY 
PROPOSITION XIX 



176. The ratio between the focal radius and the diredral dis- 
tance of any point on an ellipse is constant and is equal to the ec- 
centricity of the ellipse. 




lyCt P be any point on the ellipse, and let DR be the direc- 
trix and F the focus. Join P to F and draw PM perpendicu- 
lar to DR. 

Let ^^the eccentricity. 

We are to prove that 

PF _ 
PM ~^' 

Draw the ordinate PK. 

Let X = CK and J/ = PK. 

a = CA, 
^=CF, 
and p = FH. 
/ = FH=FA+AH. 
FA 



[i] 
[2] 



[3] Hence AH = 



AH 
FA a 






a 



ae 



by § 173. 
by § 128, [2]. 



THE ELLIPSE 119 

[4] Hence 

^ . , a — ae .a — ae , a — ae 

/ = FAH — = a — c-\ — =a — aeA, — -, 

e e e 

_ _ ae — ae^ + « — ae a{\ — ^^) 
[5] or /= ^ .= . 

[6] Also x — Q^\ — YiY^ — c-\-p—m — c—{m—p). 

The equation of the ellipse is 

[7] a'f-\-b''x'—a''b\ by §111 

Since the point P is on the ellipse, the ^ andj^' of the equa- 
tion of the ellipse may stand for the coordinates of that point. 
Hence substituting the value of x given in [6] into [7], we 
get 

[8] ay-\-b\c—{m—p)Y = cL'b\ 

[9] or «y + U'l/ — 2c{m —p) + {m—py\ — d'b' . 

b'C b~c b' 
[lol Hence r^ -\ -. 2-^ im — p) -\ tt {m — pY = b'- 

[11] But ^=i—e\ bv§i28, [5]. 

a 

[12] and c--=za-e^. by § 128, [2]. 

Substituting these values into [10], we get 

^ [13] y+^V— 2^(w— /) + (I— ^') {m~py~b\ 
[14] 

a 

[15] 
y + (?^— /)^ = — ^V+ 2 ^^{m — p^-^r e'im — PY ^ b\ 

From [11] we get 

[16] b'' — a\i~e'). 

Substituting this value of b'^ and the value of p given in 
[5] into [15], we get 

[17] y+(w — pY^=e'^ni^, 



'+b'e'~2 ^{m—p)-\- {m—pY~e\m—pY~b\ 



I20 ANALYTIC GEOMETRY 



[i8] or PK+FK = ^'HK. 

[19] Hence PF = ^"HK. by Geom. 26. 

[20] But HK = PM. by Geom. 17. 



[21] Hence PF = ^'PM, 

PF 
PM 



PF 
[22] and --—-: = e. 



Q. K. D. 



177. Corollary i . — In an ellipse the focal distance is less than 
the directral distance. 

For e is less than i. by § 129. 

Hence PF is less than PM. 

178. Corollary 2. — The distance from- the centre of an ellipse 



to the directrix is equal to 



a 



e 

For in Fig. 55 

[i] QV—c^ae, by §128, [2]. 

2 
[2] and FH=/>= . by § 176, [5]. 



[3] Hence CH =: CF+ FH = a^ + ^— ^' = — 

-^ e e 



PROPOSITION XX 

179. The equation of the ellipse when any pair of conjugate 
diameters are taken as the axes is 

in which a' and d' are the semi-conjugate diameters. 

lyCt P be any point on the ellipse, and L/M and NO be any 
two conjugate diameters. 

I^et lyM be the new axis of abscissas and NO the new axis 
of ordinates. 

Draw PS II YY' and PK II NO. 




121 



Let X = CS and jK ^ PS. 
y = CK " y = PK. 
^' = CM " /^' = CN. 
^=MCA" O'^NCA. 
We are to prove that 

is the equation of the ellipse referred to the diameters LM 
and NO. 

[r] PR= PK.sinPKR=ysin0'. by Trig, i and 22. 

[2] RS = KH = CK. sin 6 =: x' sin 0. by Trig. i. 

[3] Hence J/ = PR+RS=y sin ^' + ji;' sin 6. 

[4] RK =: PK. cos PKR = — ycos 0'. by Trig. 2 and 23. 

[5] CH = CK. cos ^ = x'cos ^. 

[6] Hence ;r = CH — RK = jc' cos -^y cos 0'. 

When XX' and YY' are taken as the axes the equation of 
the ellipse is 

[7] a-y -{- d~x^^=:a^d^. by §111. 

Substituting for the x and j/ of this equation their values 
given in [3] and [6], we get 

[8] a'ly sin^ 6' + 2xy sin 6 sin d' + x" sin^ ^] 

+ b'ly cos'O' + 2xy cos d cos O'+x" cos^ e-] = a'd\ 



t22 . ANAL YTIC GEO ME TR Y 

[9] Hence {a^ sin" 8 -\- b' cos'' Q)x^^ 
+ a^sin' 6>'+3^cos ^')y' 
+ 2(a^ sin e sin (9' + <^^ cos 8 cos (9');i:>' :== a'K-. 

h- 

[10] But tan Q tan ^' = 3-. by § 158. 

[11] Hence ^^ tan ^ = — 



tan 6^" 



[12] or a' s= — '^ -^ — ^/- bv Trig. 6, 

cos C7 Sm C7' 

[13] Hence a^ sin d sin 6^' = — b"' cos cos ^'. 

[14] Hence a~ sin ^ sin Q^ + <^' cos Q cos 0' = o. 

Substituting o for this binomial in [9] we get 

[15] (a^sin^^+<^'cos^a)y^+(«^sin^6"+^^cos^6?')y^=«^^-\ 

Since the x^ and y' of this equation stand for the coordinates 
of an}^ point on the ellipse when lyM and NO are taken as the 
axes, this equation is the equation of the ellipse referred to 
LM and NO as axes. by § 39. 

Since the point M is on the ellipse, its coordinates ^' and o 
must satisfy the equation of the ellipse. by § 40. 

Substituting these values for the x^ andjj/' of [15], we get 
[16] («-' sin^ e^-b' cos= e)a^''=d'b\ 

a-b' 



[17] Hence a"^ sin^ ^ + ^' cos^ ^ 



a" 



Since the point N is on the ellipse, its coordinates o and b^ 
must also satisfy the equation of the ellipse. 

Substituting these values for the x' and y' of [15], we get 
[18] («sin^ 6^'+^'cos^ e')b''^a'b\ 

[19] Hence a' sin^ 6^ + b^ cos' ^' = ^tt^- 

Substituting the right hand sides of [17] and [19] into [15] 
we get 

1 LA 2 /,2 

r 1 CI ^^ , CI ^^ 2,2 

[20] -^ x^' + -^y =a b . 

[21] Hence «V' + -^"•^" = ^"^". 



THE ELLIPSE 



123 



Since the old axes are no longer to be used, we may drop 
the accents over the x and>' and write the equation 

[22] ^.y + 3'v = ^''^'^ 

Q. E.D. 



PROPOSITION XXI 

180. The equation of the tangent to an ellipse when any pair 
of conjugate diameters is taken as the axes is " . 

d'^x' 

y —y — — -^A^ —^). 

in which x' and y are the coordinates of the point of tangency 
and «' and b' are the semi-conjugate diameters. 




lyCt LM and NO be two conjugate diameters. 

Let LM be taken as the X axis and NO as the Y axis. 

Let PT be tangent to the ellipse at P. 

Draw the ordinates PK and P'H. 

Let x' = CK and J/' = PK. 
y=CH " y = P'H. 
a' = CM " 3' =CN. 



124 ANALYTIC GEOMETRY 

We are to prove that 

is the equation of the tangent PT. 

The secant PP' is a straight line cutting the curve in two 
points, therefore its equation must be of the form 

[i] y—y = '^^n^,{^'—^)- by §58. 

Since the point P is on the ellipse, its coordinates x' andj^' 
must satisfy the equation of the ellipse. 

Substituting them into the equation of the ellipse referred 
to conjugate diameters, we get 

[2] a'y + r-x" = a"d'\ by § 1 79. 

Since the point P' is on the ellipse, its coordinates x" and 
y must satisfy the equation of the ellipse. 

Substituting them into the equation of the ellipse referred 
to conjugate diameters, we get 

[3] ay + d"x"' = a"d'\ by § 179. 

Proceeding as in § 134, we get 






PROPOSITION XXII 



Q. K.D. 



181. When any pair of conjugate diameters is taken as the 
axes, the equation of the chord which joiyis the points of tangency of 
two tangents drawn to an ellipse from the same poi7it without it, is 

ayy + b''xx^ = a"b'\ 

in which x' and y' are the coordinates of the point fro^n which the 
two tangents are drawn, and a' and b' are the semi-co7ijugate 
diameters. 

Let PT and P'T be two tangents drawn to the ellipse from 
the same point T, 




125 



Fig 58 

Let PP' be the chord joining their points of tangency. 
Let any two conjugate diameters LM and NO be taken as 
the axes, and draw the ordinates PK, P'H and TR. 

Let x' = CR and y' = TR. 

a' = CM " d' =CN. 

We are to prove that 

a'yy + d"xx'=a"d''\ 
is the equation of the chord PP'. 

Let x" = CK and jv" = PK. 
;*;'" = CH " y" = P'H. 

The equation of PT is 



[I] 



y —y 



V'x' 



f2 

a y 



/fV-^ •^) ' 



by § 180. 



[2] Hence a^yy'^ — a^^' = — b"xx" + b"x"\ 

[3] and a'yy"+ l>"xx" = a'y" + d"x"\ 

Since the point P is on the ellipse, its coordinates, x" and 
y", must satisfy the equation of the ellipse. 

Substituting them into the equation of the ellipse referred 
to conjugate diameters, we get 

[4] a'y'" + d"x"' = a"d' \ by § 179. 

Hence substituting a'^d'^ for the second member of [3], we 
get 



1 26 ANAL YTIC GEO ME TR Y 

[5] a'^yy^^V'xx'^ = a^H'\ 

which is the equation of PT. 

Similarly we may show that the equation of P'T is 
[6] «'>y" + /^"^;r^"' ^ ar^'^ 

Since the point T is on the tangent PT, its coordinates 
x^ andjK', must satisfy the equation of PT. 

Hence substituting them for the x and y in [5], we get 
[7] «'»" + ^''^ V = aJ'b'\ 

Similarly, since the point T is on the tangent P'T, by sub- 
stituting the coordinates of T into [6], we get 

[8] «'»'" + b^'x^x^^^ = a^'V\ 

[9] Now a^'yy' + V'xx^ = a^'b^\ 

is the equation of a straight line. by § 67. 

But the coordinates x'\ y" of the point P will satisfy this 
equation, for if they are substituted for the x and y in it, we 
get a true equation, vz2., [7]. 

Hence the straight line represented by [9] must pass 
through the point P. 

The coordinates x'", y'" of the point P' will also satisfy [9], 
for if they are substituted for the x and y in it, we get a true 
equation, vzs., [8]. 

Hence the straight line represented by [9] must also pass 
through the point P'. 

Therefore since the line represented by [9] passes through 
both points P and P', it must be the chord PP'. 

Therefore [9] must be the equation of the chord PP'. 

o. E. D. 

182. Corollary. — When the transverse axis is taken as the 
X axis and the conjugate axis as the Y axis^ the equation of the 
chord becomes 

ayy' -\- b'xx^ ^ a~b~. 

EXAMPLES 

I. From the point 10, 5, two tangents are drawn to 



THE ELLIPSE 



127 



i6jK^ + 9-^^ ^ 144. Find the slope of the chord which joins 
the two points of tangency. Ans. Slope = — f . 

2. Two tangents are drawn to 1 6 jj/^ -|- 9-^^ ^= i44 at the 
extremities of the chord y =^ — f-^ + 3- Where do the tan- 
gents meet? Ans. x = 4,jj/ = 3. 

Construct a figure showing the tangents and the chord in 
each example. 



PROPOSITION XXIII 

183. The two tangents at the extremities of any chord of an 
ellipse meet on the diameter which bisects that chord. 




Fig-. 59 

Let PT and P'T' be two tangents to the ellipse at the ex- 
tremities of the chord PP', and let LM be the diameter which 
bisects that chord. 

We are to prove that PT and P'T' meet on the diameter 
LM. 

Let NO be the diameter which is conjugate to LM. 
Let LM be taken as the X axis and NO as the Y axis. 

Let «' = CM and V = CN. 
y = CK. 
Now if R be the point where PT cuts the axis LM, then 
by § 180 and § 45 



128 



ANALYTIC GEOMETRY 



[I] 



CR = 



a 



X' 



Similarly, if R' be the point where *P'T' cuts the axis LM, 
[2] then CR' 

[3] Hence CR == CR'. byGeom. i. 

Therefore the points R and R' coincide with each other and 
the two tangents meet the diameter at the same point. 

Q. E.D. 



CR'. 



PROPOSITION XXIV 

183. If two tangents be drawn at the extremities of a?iy 

focal chord', 

(i) the two tangents will meet on the directrix, and 

(2) the line joining the i?itersection of the two tangents to 

the focus will be perpendicular to the focal chord. 













Y 






D 












\ 








P_-= 


_ — — 




-r^ 


B "^■^--v 




\ 


^ ' 


■ — -^ 


/^ 


/ 




\ 


T-^ 


^ 


5 


J 


/ 


C 


\ 




K 


\ 








J 




D' 




\ 




BL--^ 


y^ 



Fig. 60 

Let PT and P'T' be two tangents drawn to the ellipse at 
the extremities of the focal chord PP'. Let R be the intersec- 
tion of the two tangents, and DD' the directrix. 

We are to prove 

(i) that R will be on the directrix DD'. 

(2) that RF will be the perpendicular to PP'. 



THE ELLIPSE 129 

Ivet ;c'= CK andy= RK. 

lyet X and^ be the coordinates of any point on the chord 
PP'. 

The equation of the chord PP' is 

[i] a^yy' + b'-xx' = a'b\ by § 182. 

in which the x' and j/' are the coordinates of the point R. 

[2] CF = «^. by § 128; [2]. 

Since the point F is on the chord PP', its coordinates ;«; = «^ 
andjj/ = o must satisfy the equation of that chord. 

Substituting these values of x andjv into [i], we get 
[3] b'^aex' =^arb'. 

[4! Hence x' = — . 

But — is the distance from the center to the directrix. 
e 

by § 178. 

Hence R, the intersection of the two tangents, must be on 
the directrix. 

Q. E.D. 

Again, since RF is a straight line passing through the two 
fixed points R and F, its equation must be of the form 

[5] y-^=$Ei!(^'-^)- by §58. 

In [5] let ^" and jj/" stand for the coordinates of the point 
F, and x^ and y' stand for the coordinates of the point R. 

Then y = ae, by § 128, [2]. 

andj>/" = o. 

Substituting these values of ;»:" and y" into [5], we get 

[6] y_^^_^:LZl_(y_^). 

ae — X 

But ■^'=T by [4]. 



I30 ANALYTIC GEOMETRY 

Substituting this value of x^ into [6] , we get 

[7] y_^::^_^>:L(y_^). 

[8] Hence y — y= ^/-^ ,A x' — x) , 
which is the equation of RF. 



[9] By [4] X 



Substituting this value of x' into [i], we get 



a 



[10] a'yy = —b''^x-\-a''b\ 



e 



[11] or a^ey^y^ — y^ ax -\- a^ b^ e . 

b"" 

ri2l Hence r = r(-^ — ^^)- 

aey 

[13] But ^■''^a^Ci— ^^). by §128, [5]. 

[141 Hence y = — -. (x — ae), 

aey 

which is the equation of the chord PP'. 

lyCt s = the slope of the line RF, 
and 5'= '' " ** '' " PP'. 



From [8] we get 




firi c- ""'y' 


by §53- 


^'5J '-a'{i e') 


From [14] we get 




^ -■ aey 


by § 53- 


aev 
"17] Hence 55'=-^- — - — ^ X — 


aey 


[18] and I + ^i-' = 0. 




Therefore RF and PP' are perpendicular to each other. 




by § 62. 




Q. E. D. 



THE ELLIPSE 



i^^i 



PROPOSITION XXV 



184. The locus of the intersection of two tangents to an ellipse 
which are perpendicular to each other ^ is a circle whose center is 
the center of the ellipse. 




Fig. 61 

Let PT and P'T be the two tangents to an ellipse at the 
points P and P'. Let them be perpendicular to each other at 
the point T. 

Let P and P' move along the ellipse, but so that PTP' shall 
always be a right angle. 

We are to prove that the locus of the point T will be a cir- 
cle whose center is at the origin. 

Let x = CK andy = PK. 

The equation of the tangent PT is 



[i] 



y —y 



ay 



(x' — x). 



by § 134. 



[2] Hence ay'jy = ay^ -\- d^x'^ — d^x'x. 

Now since the point P is on the ellipse, its coordinates must 
satisfy the equation of the ellipse. by § 39. 

[3] Hence a^ + 5'x" = a'd't by § 1 1 1 . 

Substituting this value of ay^ + d^x'''^ into [2], we get 

[4] ^yy =^ cL'b^ — b'x^x. 



132 ANALYTIC GEOMETRY 

U" b~x' 
[5] hence _^:^___^. 

[6] But -^=^i;=^^^^^whichby[3] 

[7] Hence ^=V^^ + .^-g;=V^^+.^(^;)\ 
Now substituting this value of — j into [5], we get 

[8] ^=-5f;-+V*'+'''(5fy- 

_ b'x' 

a J/ 

then [8] becomes 

[9] y ^^ sx -\" V b~ -\- a^ s^ ^ 

which is the equation of any tangent PT to an ellipse. 
[10] Let y—s'x^ Vb' + a's'\ 

be the equation of the tangent P'T. 

Then since FT and P'T are JL to each other 

[11] iH-^5' = o. by §62. 

[12! Hence / = — - — . 

. s 

Substituting this value of s' into [10], we get 

[13] ,i' = -™+\/*' + ii 

o S 

which is the equation of P'T. 

By transposition [9] and [13] become 

[14] y — sx = Vb'+a's\ 

[15] y + ^^^^' + ^- 






THE ELLIPSE 133 

From [14] we get by squaring 

[16] y' — 2syx + s^x- = Z>^ -|- «V, the equation of PT. 

[17] From [15] y+2^ + ^ = *=+^. 

Clearing [17] of fractions, we get 

[18] ry + 2sxy + x'= ^V + a\ the equation of P'T. 

Since the point T is on both tangents its coordinates must 
satisfy [16 J and [18]. Hence we will let the x and_y of [16] 
and [18] be the coordinates of T. Then these equations are 
simultaneous and may be combined. 

Adding [16] and [18], we get 

[19] (1 + s')y+{i +s')x' = a\i +s')+d\L + s'). 
[20] Hence y -{- x^ =z a~ -\- d^, 

in which x andjv are the coordinates of the point T. 

Hence [20] is the equation of the locus traced out by the 
point T when the points P and P' move along the ellipse. 

by § 39. 
Let a' + d' — r\ 

[21] Then [20] becomes x^ -\- y z= r^^ 

which is the equation of a circle whose center is at the 
origin. by § 113. 

Therefore the locus traced out by the point T is a circle 
whose center is the center of the ellipse. 

Q. K.D. 



134 



ANALYTIC GEOMETRY 



PROPOSITION XXVI 

185. If aiiy chord of an ellipse pass through a fixed point and 
tangents be drawn at its extremities, and if the chord be made to 
revolve about the fixed point as a pivot, then the locus of the in- 
tersection of the two tangents will be a straight line whose equa- 
tion is 

ayy + b^'xx^ — a''b'\ 

in which x' and y are the coordinates of the fixed point about 
ivhich the chord revolves, anda^ and b^ are the semi-conjugate diam- 
eters which are taken as axes. 




Figf. 62 



L<et PP' be the chord passing through the fixed point R and 
let PT and P'T be the two tangents drawn to the ellipse at its 
extremities. 

I^et lyM and NO be two conjugate diameters taken as axes. 

lyet x' = CH andy = RH, 

Let PP- revolve about R as a pivot. 

We are to prove that the locus traced out by T is a straight 
line and that its equation is 

a'yy + b'^xx^^ = a"'b'\ 
Ivet ^" = CK andy ^ TK. 



THE ELLIPSE 135 

The equation of the chord PP' is 
[i] «''_>y+<5'''A-;^:" = «''^'^ by § 181. 

Since the point R is on this chord, its coordinates must sat- 
isfy the equation of this chord. by § 40. 

Hence substituting Jt:' and _>/' for the ;»r and jf in [i], we get 

[2] a'>y' + ^'vy zz:«'^^'^ 

Now as PP' revolves about R, T will trace out a locus. 
Moreover [2] will be true for the coordinates of the point T 
wherever it may be as it traces out this locus. Hence the jr" 
andy of [2] stand for the coordinates of every point on the 
locus traced out by T. Therefore [2] must be the equation 
of that locus. by § 39. 

This locus must also be a straight line. by § 67. 

Since T is any point on the locus traced out by the inter- 
section of the tangents, we may drop the accent marks from 
its coordinates x" andy and write them x and y. Then [2] 
may be written 

[3] a'yy+d"xx'=a"d'\ 

Therefore the locus of T is a straight line and its equation 
is [3]- 

Q. E.D. 

186. Corollary. — When the axes of the ellipse are taken as the 
axes of coordinates, the equation of the locus of the i7iter section 
of the tangents becomes 

ayy' -f b'xx' — a:'b\ 

187. Supplemental Chords. — Two chords drawn from the 
same point on an ellipse to the extremities of any diameter 
are called supplemental chords. 



136 



ANALYTIC GEOMETRY 



PROPOSITION XXVII 

188. If a chord be parallel to any diameter of an ellipse, the 
supplemental chord will be parallel to the conjugate of that diam- 
eter. 




Fig. 63 



Let MC and MC be two supplemental chords drawn to the 
extremities of the diameter CC 

Let DD' and EE' be two conjugate diameters and let 

MC II EE'. 

We are to prove that 







MC II DD^ 




[I] 




C'0_C'K 
C'C~"C'M* 


by Geom/ 23. 


[2] 


But 


CO , 

cc~^- 


by § 163. 


[3] 


Hence 


CK 1 
CM~2' 




[4] 


Hence 


CK= JC'M. 





That is, the diameter EE' II MC bisects a system of chords 
II MC. But by hypothesis EE' and DD' are conjugate, and 
therefore EE' bisects a system of chords parallel to DD'. 

by § 157- 
Therefore MC i| DD'. by Geom. 10. 

Q. K. D. 



THE ELLIPSE 



137 



Polar Equations of the Ellipse 

PROPOSITION XXVIII 

189. When the right hand focus is taken as the pole, the polar 
equation of the ellipse is 

a(i — e^) 
I + ^ COS Q 

in which a is the semi-transverse axis of the ellipse, e is its eccen- 
t?^icity, r the radius vector of a7iy point on the ellipse, and d the 
vectorial angle. 



t 







N. p 


D 




/\ 


i 


c 


A \k 


E 






I K 


0' 



Fig. 64 

Let F be the pole, XX' the initial line and DD' the direc- 
trix. 

Let P be any point on the ellipse. Draw PD _L the direc- 
trix and PK_L XX'. 

Let ^ = XPFX, r = FP,and ^ = the eccentricity. 

We are to prove that 

a{i — e'^) 





I -\- e cos 6 




is the polar equation 


of the ellipse. 




]i" 


PF = ^PD. 


by s 176 


^2] Hence 


PF = ^(EF — FK), 




is] or 


PF = ^EF — ^FK. 




>] But 


EF-"^^ ''\ 


by §176, [5; 



138 ANALYTIC GEOMETRY 

[5] and FK = r cos d. by Trig. 2. 

[6] Hence by [3] PF = r=:<2(i — e^) — eraosQ, 
[7] and r(i -j-^cos ^) — tz(i — e^) . 

Mail — e-) 
Hence r — -^, ^. 

^ -\- e cos C7 

Q. E.D. 

1 90 . Corollary i . — The polar equation of the ellipse will become 

a(i — e') 

r —^ 

I — e cos d 

when the left hand focus is the pole. 

191. Corollary 2. — The polar equation of the ellipse may be 
writteji 

p 



r = 



2{\-\- e cosdy 
in ivhich p is the param-eter . 

[i] For (i_^')=il. by §128, [5]. 

Substituting into [8], § 189, we get 

b' 1 



v^l 










a I -f- ^ ( 


:os d' 


Let/ 


be the 


parameter. 






j! 


Then 






p 

2 


: b:: b: 


a. 


'a. 


Hence 








p _b^ 
2 a 


• 


Substituting 


into 


"2" 


we get 




^5; 






r- 


1 


P 




(I +^cos^)' 



by § 125. 



Q. K.D. 



EXAMPLES 



I. What is the value of r when = o? What line in Fig. 
64 does r then represent ? Ans. rz=z a — ae. 

r — FA. 



THE ELLIPSE 



139 



2. What is the value of r when d = 180° ? What line in 
Fig. 64 does r then represent. Ans. r^=- a-\- ae. 

r— FA'. 

3. What is the value of Q when r is drawn to the extremity 
of the conjugate axis? 

Ans. ^ :=: cos""' ( — (?) . 

4. The eccentricity of an ellipse is \. What is the value of 
B when r ^^\a'> Ans. ^ = o. 



PROPOSITION XXIX 

192. When the pole is at the center the polar equation of the 
ellipse is 



f^ = 



i—e'cos'd' 



in which b is the semi-conjugate axis of the ellipse^ e its ecceii- 
t7'icity, r the radius vector of a7iy point on the ellipse, and 6 the 
vectorial angle. 




Fig. 65 

I^et P be any point on the ellipse and drawuts ordinate PK. 

^=^PCK and /-^CP, 
«= CA " <^ = CB. 

We are to prove that 

b'' 



r' = 



I — e^ cos^ 6' 



I40 ANALYTIC GEOMETRY 

Let jr = CK and jK = PK. 
[i] X = r cos Q . by Trig. 2. 

[2] y = r sin 0. by Trig. i. 

When YY' and XX' are taken as the axes of coordinates, 
the equation of the ellipse is 

[3] ay-\-b'x' = a'h\ by §111. 

Substituting the values of x and y given in [i] and [2] , 
into [3] , we get 

[4] a'r^ sin' 8 + b'r cos' Q = aH\ 

[5] Hence ^ ^ ^^^^ e + ^^o^S' 

[6] But sin' d—\ — cos' Q. by Trig. 5. 

. -, __ „ ^'^' 



L/J 


J-J-CIH^C 


r — 


~ a' («' 3')cos'e' 


>; 


and 


r' 


^' 


I 5 — cos u 


L9] 


But 




a' b' 
a 


[10] 


Hence 




v — 


I ^ COS u 



by § 128, [4]. 



Q.E.D. 

193. Corollary . — When the pole is at the center the polar equa- 
tion of the ellipse may be written 

a\i — e') 
r- = — 



,2 „„2 0- 



I — e cos 

3' 

[i] For -^=i--e\ by §128, [5]. 

[2] Hence ^' = a'(i — ^'). 

Substituting this value of b^ into [10] § 192, we get 

[3] r'= ^ 



I— e' cos' 6' 



THE ELLIPSE 141 

EXAMPI^ES 

1. The center being the pole, what is the value of r when 
0=0? Ans. r ^ a. 

2. What is the value of r when d = 90° ? Ans. r^=d. 

3. In an ellipse whose eccentricity is ^, what is the value 
of ?" when =: 60° ? Ans. ^=0.89^. 

4. What is the value of r in the equilateral ellipse? 

Ans. r^ b. 

5 . Two tangents are drawn from a point to a circle ; re- 
quired the equation of the chord of contact in the following : 

(i) From (4, 2) to ^' +jK^ = 9. Ans. 4^1; +2^' = 9. 

(2) From (<2, b^ X.o x^ ■\- y'^ ^r^ c^ . Ans. ax-\-by^^c^. 

6. Find the equation of a circle through (4, o), (o, 4), 
(6,4). Ans. x^ -\- y'^ — ^x — 6_>/ + 8 = o. 

Suggestion. — ^Join (4, o) to (o, 4) and to (6, 4) by straight lines ; then 
erect perpendiculars at the middle points of these two lines ; their inter- 
section will be the center of the circle, and the distance from this cen- 
ter to any one of the points will be the radius. Then make use of 
§115. 

7. Find the equation of a circle through (o, o), ( — 8«, o), 
(o, 6a). Ans. x^ -{- y -\- Sax — 6ajK = o. 

8. Find the equation of a circle through (10, 4), (17, — 3), 
and radius =13. 

9. Find the equation of a circle touching each axis at a 
distance of 4 units from the origin. 

Ans. x^-\-y — 8x — 8y -\- 16 — o. 

10. Find the equation of a circle through (5, 6), and hav- 
ing its center at the intersection of jf= jx — 3 ; 4jj/ — 3x= 13. 

Ans. (x — lY -\- iy — 4)^=20. 

11. What must be the value of ^ in order that the line 
y ^ sx — 4 may touch the circle ;t^ + jk^ = 2 ? 

Ans. ^ = ± |/ 7. 

12. Required the equation of a tangent to the ellipse 



X y 

f- ^ := I, whose inclination to the X axis is 45°. 



142 ANALYTIC GEOMETRY 

13. If 3j = 5;f is a diameter of \-^^ = i, what is the 

4 9 

equation of the conjugate diameter ? Ans. 2oy-\- 29:^= o. 

2 2 

SV V 

14. Required the area of \- ^^—- =1 i. Ans. 27ri/io. 

4 ro ^ 

15. The extremities of a line of constant length slide along 
the coordinate axes. Required the locus traced by any point 
of the line. 

16. If from the extremitj^ of any diameter straight lines be 
drawn to the foci, prove that their product is equal to the 
square of half the conjugate diameter. 

17. Find the equation of a diameter parallel to the normal 
drawn to the ellipse at {x, y) , the semi-axes being a and b. 

18. Write the equations of diameters conjugate to the lines 
X — y = o ; X -\- y ^=^ o \ ax = by ; ay =: bx. 

Ans. b'^x + a^y =: o ; b^x — a'^y r= o ; 
a^y -|- b'^x = o ; bx -\- ay ::^ o. 

19. The center of an ellipse is at (4, 7), the major and 
minor axes are 14 and 8. Required its equation, the axes 
being parallel to the axes of coordinates. 

20. Prove that the length of a line drawn from the center to 
a tangent and parallel to either focal radius of the point of 
contact is equal to the semi-major axis. 

21. The minor axis =12, the double focal ordinate = 5, re- 
quired the equation of the ellipse, the origin being at the left 
hand vertex. 

Ans. ^^+y 1=5^. 
144 



CHAPTER X 

The Hyperbola 

194. The Hyperbola. — The hyperbola is the locus of a 
point moving in a plane in such a way that the difference be- 
tween its distances from two fixed points in the plane is con- 
stant. 




I^et F and F' be the two fixed points in the plane. Let P 
be a point moving in this plane in such a way that PF' — PF 
is constant. 

Then the line PAR traced out by P is one branch of an h}^- 
perbola. 

The line P'A'R' traced out by a point P' moving in the 
same way as P is another branch of the hyperbola. 
The whole locus PARP'A'R' is an hyperbola. 



PROBLEM 

195. To draw an hyperbola. 

Let F'H (Fig. 67) represent a ruler which maj^ be moved 
about the point F' as a pivot. 

Let /^the length of the ruler and a = any constant. 

Take an inelastic thread whose length is / — 2a and fasten 
one end of it at F and the other at H'. Place a pencil point 



144 ANALYTIC GEOMETRY 




Fig. 67 

against the thread so as to form a loop at some point P on the 
ed^e of the ruler. Now, keeping the thread stretched and the 
pencil point against the edge of the ruler, turn the ruler about 
the point F'. 

As the point P moves F'P and FP are both increased or 
both decreased by the same amount. Hence F'P — PF re- 
mains constant, and therefore the locus traced out by P will 
be one branch of an hyperbola. 

The other branch of the hyperbola is drawn by taking the 
point P as the pivot about which the ruler is turned. 

It will be shown hereafter that in Fig. 67 

196. Corollary. — // is obvious that the locus drawn hi this way 
must cut the line F' Fin two points A and A' . 

197. The Foci.— -The two fixed points are called th^foci. 

198. The Focal Radii. — The distances from the foci to any 
point on the hyperbola are called the focal radii of that point. 

199. The Vertices. — The points in which the hyperbola 
cuts the straight line passing through the foci are called the 
vertices of the hyperbola. 

200. The Transverse Axis. — The line which joins the ver- 
tices is called the transverse axis, 

201. The Center. — The middle of the transverse axis is 
called the center oi the hyperbola. 



THE HYPERBOLA 



145 



202. The Conjugate Axis. — A straight line drawn through 
the center perpendicular to the transverse axis, bisected by it 
and equal to twice the square root of the difference between 
the square of the distance from the focus to the center and 
the square of the semi-transverse axis is called the conjugate 
axis. 



PROPOSITION I 

203. The difference between the focal radii of any point on the 
hyperbola is equal to the tra^isverse axis. 




Fig". 68 



Let P be any point on the hyperbola, r and r' its focal radii, 
and AA' its transverse axis. 

We are to prove that 

r' — r=^M. 

Let the point P move along the branch PAR of the hyper- 
bola. When it reaches A we have 



[i] 



r' 



F'A — FA. 



by § 195, 



Now let P' move along the branch P'A'R' until it reaches 



A^ 



[2] Then r' — r= FA' — F'A'. 

From [i] and [2] we get 

[3] F'A — FA = FA'— F'A'. 

[4] Now F'A = F'A' + A'A, 

[5] and FA' = FA + AA'. 



146 



ANALYTIC GEOMETRY 



Substituting these values into [3] we get 

[6] FA' + A' A — FA = FA + A'A — FA' ; 

[7] hence 2F'A' = 2FA, 

[8] and FA' = FA. 

Substituting the right hand side of [4] into [i], we get 

[9] r' — r = FA' + A'A — FA. 

But from [8] we get 

[10] FA' — FA = o. 

Therefore [9] becomes 

[11] r'— r=A'A. 

Q. E. D. 



PROPOSITION II 
204. The equation of an hyperbola is 

aY — ^"^' = — a'd\ 

in which a is the semi-transverse axis, bthe semi-conjugate axis, 
and X and y the coordinates of any point on the hyperbola. 




Fig. 69 

Let BB' be the conjugate and A A' the transverse axis of 
the hj'perbola. 

Let AA' and BB' produced be the axes of coordinates. 

Draw PK 11 YY'. 



THE HYPERBOLA 



147 



Ivet ^ = CK and 7 = PK, 

We are to prove that 
[i] a/y^ — b''x^ — —d'b'' 

is the equation of the hyperbola. 

I.et r = PF, r' = PF', and c = CF. 



[2' 




r'^ = F'K + PK. 


by Geom. 26. 


[3] 


But 


V"^ = x+c. 




'a. 


Hence 
or 


y^^^x + cY+f, 




[5] 


r' = y{x^cy^y\ 




[6] 




r' = FK + PK. 


by Geom. 26. 


[7] 


But 


FK = x — c. 




>; 


Hence 
or 


r'= {x-cy+y\ 




's. 


r= y/{x — c)' + y\ 




10' 


But 


y — r = 2a. 


by § 203. 



[11] Hence ^ {x + cY -^ f ~ -y^ {x —cY ^ y- = 2a. 
Clearing this equation of radicals as in § i it, [ii], we get 
[12] ay — (r — a')x'' ^—a\c'—a^). 

[13] But 
[14] Hence 



by § 202. 



b — y^c' — a\ 

b' ^r — a\ 

Substituting this value of c^ — a^into [12] , we get 

[15] ay — b'x' = —a'b\ 

Since in [15] the x and y stand for the coordinates of any 

point on the h5^perbola, that equation must be the equation of 

the hyperbola. 

Q. K. D. 

205. Corollary I. — Since i?i the ellipse b'^ stands for a^ — c^ 
but in the hyperbola it stands for c' — a^ z=. — {a^ — ^^), any 
equation of the ellipse may be changed into the corresponding 
equation of the hyperbola by substituting — b^ for b"^ . 



148 ANALYTIC GEOMETRY 

206. Corollary 2, — The distance fro in the center to the focus is 
equal to the distance from the vertex to the extremity of the con- 
jugate axis. 



or 




If ^^ a\ 


by § 202. 


'2_ 


Hence 


r = a' + ^. 




[3] 


But 


a=+3^ = AB. 


by Geom. 26. 


[4] 


Hence 


c- = AB, 




;5; 


and 


CF = AB. 





207. The Equilateral Hyperbola. — The equilateral hyper- 
bola is that hyperbola whose axes are equal to each other. 

208. Corollary. — The equation of the equilateral hyperbola is 

y — X =: — a i 
in which a is the semi-transverse axis. 
The equation of any hyperbola is 

[i] d'y' — b^x'^^ — a'b'. by § 204. 

But in the equilateral hyperbola 

[2] «^ = <^^ by § 207. 

Hence dividing [i] by [2] member by member, we get 

[3] f — x'^—a:' 

Compare with § 113. 



PROPOSITION III 

209. If to an hyperbola and the equilateral hyperbola which 
has the same transverse axis^ ordinates be drawn to the same 
point 071 the transverse axis, then the ordinate of the first hyper- 
bola will be to the ordinate of the equilateral hyyerbola as the 
conjugate axis is to the tra?isverse axis. 

Let PAM be any hyperbola, A A' its transverse axis, and 
BB' its conjugate axis. 



,2 




THE HYPERBOLA 

Y 






149 




Y' 

Fig. 70 



Iret P'AL be the equilateral hyperbola having the same 
transverse axis. 

Draw the ordinates P'K and PK to the same point K. 

Let ;i: ^ CK, J/ = PK and / = P'K. 
Let « = CA and ^ = CB. 

We are to prove that 

PK : P'K :: BB' : AA'. 

Since P is on the hyperbola PAM, its coordinates x and y 
must satisfy the equation of that hyperbola. by § 40. 

Hence, letting x and y of the equation of the hyperbola 
stand for the coordinates of the point P, we get 

[i] <2y — b'^x'^ = — a'U^, by § 204. 



[2] or 



y = Tr(^^-^') 



a 



Since the point P' is on the equilateral hyperbola PAL, 
its coordinates x and_y' must satisfy the equation of the equi- 
lateral hyperbola. by § 40. 

Hence substituting x and >'for the x and _>/ of that equation, 
we get 

[3] y^ =. x'' — a\ by §208. 

Dividing the members of [2] by the corresponding mem- 
bers of [3] , we get 



ISO 



[4] 

[5] Hence 
[6] or 



ANALYTIC GEOMETRY 

y~ a ~AA" 
J :>' :: BB' : AA'. 



Q. K. D. 



210. Conjugate Hyperbolas. — Two hyperbolas are conju- 
gate to each other when the transverse axis of each is the con- 
jugate axis of the other. 



PROPOSITION IV 



211. The equation of the conjugate to any hyperbola is 



aY—b'x'' = a'b\ 



in which b is the semi-transverse and a the semi-conjugate axis 
of the conjugate hyperbola. 




Let PBB'M be conjugate to the hyperbola RAA'S. 

Let « = CA and b = CB. 
Let P be any point on the conjugate hyperbola PBB'M. 
We are to prove that the equation of the hyperbola PBB'M is 

ay — b'x' = a'b\ 



THE HYPERBOLA 151 

If for the conjugate hyperbola PBB'M we take YY' for the 
X axis, and XX' for the Y axis, then since the point P is on 
the hyperbola PBB'M, its coordinates x = PH and y = PK 
must satisfy the equation of the hyperbola. by § 40. 

But for the conjugate hyperbola, d is the semi-transverse 
axis, and a is the semi-conjugate axis. by § 210. 

Hence substituting PH for x, PK for jf, ^ for a and a for d 
into the equation of the hyperbola, we get 

[i] d'FK~a-YH = — a'd\ by § 204. 

But for the hyperbola RAA'S and in all other theorems, we 
have taken XX' for the X axis and YY' for the Y axis. 
Hence for the sake of uniformity, we will do the same for the 
present theorem. 

Then PK = CH will be represented by x and PH will be 
represented byjK. 

Substituting x for PK and y for PH in [i], we get 

[2] d'x' — ay' = — a'd\ 

[3] Hence aY — d'x' - a'b\ 

Now since in [3] x 2,w^ y are the coordinates of P, and 
P is any point on the hyperbola PBB'M, [3] must be the equa- 
tion of that hyperbola. by § 39. 

Q. E. D. 

Corollary, — The equations of any hyperbola and its conjugate 
differ only in the signs of their absolute terms. 



PROPOSITION V 

212, The squares of the ordinate s of any two points on an hy- 
perbola are to each other as the products of the segments which 
they make on the transverse axis. 

lyCt P and P' be any two points on an hyperbola, and let 
PK and P'H be their ordinates. 

Let x^ = CK and / = PK. 
;c"^CH " y' = P'H. 



152 




ANALYTIC GEOMETRY 

Y 



+B 
C 



P> 




F K H 



-Mil 



B' 



Y' 

Fig. 72 

We are to prove that 

y : >"^ : : A'K.KA : A'H HA. 

Leta = CA=CA'. 

[i] A'K = a^x' and KA =- ^' — a. 

[2] A'H = a + y and YLK = x'' — a. 

Since the point P is on the hyperbola, its coordinates jr' 
andjj/' nlust satisfy the equation of the hj-perbola. by § 40. 

Substituting x^ andji^' for the x and y of that equation, we 
get 

[3] a^y' — b'x'' = — d'b\ by § 204. 

Similarly, since the point P'is also on the hyperbola, we get 

[4] d'y"- — b'x'" = — db\ 

By transposing and factoring, [3] becomes 

[5] ay''=b'{x''-d), 

and [4] becomes 

[6] ay'^b^x'^ — d). 

Dividing the members of [5] by the corresponding mem- 
bers of [6] , we get 



[7] 



y" _x'^ 

y/2 — ^fri 



a' 



a 



THE HYPERBOLA 153 

[8] Hence y''' -y'" v. {x' + a){x'—a) -{x''-^ a){x"—a) , 
[9] or y- : y'" ': : A'K.KA : A'H.HA. 

O. E. D. 

213. Corollary i . — Ordinates at equal dista^ices from the cen- 
ter are equal. 

214. Corollary 2. — The hyperbola is symmetrical with respect 
to both axes. 

215. Corollary ^. — If the ordinates of any two points on an 
hyperbola be at equal distances from the ceyiter^ the points will be 
equally distant from the adjacent foci. 

That is if CH = CH', 

then will FT = FP'. 

216. The Parameter. — The param.eter of an hyperbola is 
the double ordinate which passes through the focus. 

217. Corollary. — The parameter is a third proportional to the 
transverse and conjugate axis. 

Since the point R is on the hyperbola, its coordinates CF* 
and RF must satisfy the equation of the hyperbola, by § 40. 
Hence substituting CF for ^ and RF for y in that equation, 



we get 










]i] 




«-.RF b\Q,V = a'UK 




by § 204. 


'2 


But 


CF := a^ + b\ 




by § 202. 


'.i. 


Hence 


^\RF b\a'^b')^ a'b' 


) 




'a. 


or 


a'.'R.^—a'b' — b'^— a'b\ 






[5] 


Hence 


a^RF = b\ 






'6\ 




^.RF = b\ 






.7. 




2a.2RF = ^b' . - 






>; 




2RF : 2b : \ 2b : 2a. 


by 


Geom. 56. 


[9] 


or 


RS : BB' :: BB' : AA'. 




by § 213. 



154 ANALYTIC GEOMETRY 

218, The Eccentricity. — The eccentricity of an hyperbola 
is the quotient of the distance from the focus to the center 
by the semi-transverse axis. 

lyCt e = the eccentricity. 

In Fig. 72 let ^ = CF and a ■=, CA. 

[I] Then ^=1^ = ^. 

[2] and c = ae. 

[3] But az-^-h' — c\ by § 202. 

[4 J Hence ^ — = e% 



[5] and 



a 



219. Corollary. — The eccentricity of an hyperbola is greater 
than I . 

For in [i], § 218, CF is greater than CA. 



KXAMPLES 

1. What are the semi-axes and the eccentricities of the fol- 
lowing hyperbolas ? 

25j|/^ — 16^^ = — 400, 

3y— 2^' = 12, 

and y- :=: ni, 

4 

2. The equation of an hyperbola is i6j/^ — gx'^ ■:=: — 144. 
What is the distance of the focus from the center, and the 
distance of the focus from each of the vertices ? 

3. The divStance from the vertex of an hyperbola to the end 
of the conjugate axis is 5, and its semi-transverse axis is 4. 
What is the eccentricity and^he equation of the hyperbola ? 

4. Find the eccentricity of an equilateral hyperbola. 



THE HYPERBOLA 



155 



PROPOSITION VI 

220. If r' be the longer and r the shorter focal radius of any 
point on the hyperbola, then 

r' m ex -\- a, 
and r = ex — a, 

in which e is the eccentricity and a the semi-transverse axis of the 
hyperbola. 




In Fig. 73 

let r' = F'P and r = FP, 
^ = CK " jj/ = PK, 
and ^=CF " « = CA. 

We are to prove that 

r" ^=ex -\- a, 
and r^ ex — a. 

[i] r''=y-+{x + cy. 

[2] Hence r'^ n^ y~ -\- x^ -\- 2cx -\- <f^ 

[3] But ^y — b'x'' = —a'b\ 

[4] Hence / ^ K {x' — a'^ . 

a~ 

Substituting^ this value of^ into [2], we get 



byGeom. 26, 
by § 204, 



[5] 



b' 



a 



{x^ — a^) -\- x^ -\- 2CX -\-c^ 



156 



[6 
[7 

[8 
[9 



ANALYTIC GEOMETRY 



But c'-b'-^a\ by§202. 

}Ience r'- = -^ {x^ — a') -\- x' ~{- 2cx -\- a^ -\- b^ , ^ 

b' + x'+2cx^a''+b\ 

by §2i8, [2]. 
x^ + 2aex + a^ , 



or 

Now 
Hence r' 



r" = 



a' 






a 



[11] or r'^=^'^^+ 2^^Jt: + ^' = (^;r + <3;)^ by§2i8, [4]. 
[12] Hence r^ := ex -\- a. 

Similarly it may be shown that 

[13] r :^ ex — a. 

Q. B. D. 

PROPOSITION VII 
221. The equation of the tangent to an hyperbola is 

in which x' and y' are the coordinates of the point of tangency, 
and a and b are the semi-axes. 




Let PT be a tangent to the hyperbola at the point P. 
Let x' = CK and 7' ^ PK. 
^ = CA " b ^ CB. 



THE HYPERBOLA 157 

We are to prove that » 

•^ -^ a' y 

Let PM be a secant cutting the hyperbola at the two points 
P and P'. 

Let ^"= CH and /' = P'H. 

Since the secant is a straight line passing through two 
fixed points, its equation must be of the form 

[i] y— ->'=-5E|;'^-^'-^)- by §58. 

Since P is on the hyperbola, its coordinates x^ andy must 
satisfy the equation of the hyperbola. by § 40. 

Hence substituting ;r' and jv' for the ^ andjK of that equation, 
we get 

[2] ay'—d"-x" ^—a'b\ by § 204. 

Similarly, since the point P' is on the hyperbola, its coor- 
dinates Jtr" and J/" must also satisfy the equation of the hyper- 
bola, by § 40. 

Hence substituting ^" and y'^ for the x and y of that equa- 
tion, we get 

[3] a>"^ — ^^^"^ = — a'b\ by § 204. 

Subtracting the members of [2] from the corresponding 
members of [3], we get 

[4] aHy''' —y") — d'(x"' — x") = o, 

[5] or ^'(y— y) = d'{x"' — x"). 

[6] Hence a'iy" —y){y -\-y') = d"-{x" —x') {x'-\-x") , 
y —y _ b' x' + x" 



[7] and 



x' a' y H-y 



Substituting this value oi"^, =^ into [i], we get 

.[8] y_^=i; !;+£;;(..-.). 

Now let the point P' move along the hyperbola towards P. 



t58 ANAL YTIC GEO ME TR Y 

Then the secant will revolve about the point P as a pivot and 
will continually approach the tangent PT, and when P' reaches 
P it will coincide with the tangent. 

When P' reaches the point P we will have 
[9] x" = x' and y" ^ y , 

and the fraction in [8] becomes 

b^ 2x' _b'x' 
'-''^-' a' 2y'~ a'y" 

and [8] becomes 

b^ x^ 
[11] y —y= -^— (^x' —x). 

ay 

Now the X and J/ of [8] stand for the coordinates of every 
point on the secant in every position which it takes as it re- 
volves about P. Hence they stand for the coordinates of 
every point on it when it coincides with the tangent. 

But when the secant coincides with the tangent, [8] takes 
the form of [11]. Hence the x and y of [11] stand for the co- 
ordinates of every point on the tangent. 

Therefore [11] is the equation of the tangent. by § 39. 

Q. E.D. 

b"^ x' 

222. Co7'ollary. — The fraction —^ — -, is the slope of the tangent. 

For proof compare § 135. 

223. The Subtangent. — The subtangent is the distance 
measured along the X axis from the ordinate of the point of 
tangency to the tangent. 



Corollary i. — The length of the subtangent is 



x'^^ — a' 



X' 

For proof compare § 138. 

224. Corollary 2. — If different hyperbolas have the same 
trafisverse axis and ordinates be drawn to each from the same 
point on the transverse axis, then all the subtangents will be 
equal to each other. 

Compare § 140. 



THE HYPERBOLA 



159 



Draw a figure showing all the lines referred to in this cor- 
ollary. 

225. The Normal, — The ?2^r;;2<2/ to an hyperbola is a straight 
line perpendicular to the tangent at the point of tangency. 



PROPOSITION VIII 
226. The equation of the normal to an hyperbola is 

y 






in which x' and y are the coordinates of the point of tangency, 
and a and b are the semi-axes. 




Let PN be the normal and PM the tangent to the hyper- 
bola at the point P. 

Let x' = CK andy = PK, 

and s' ^ the slope of the tangent PM. 

We are to prove that the equation of PN is 



y —y — 



ay 



b^x 



,{x'—x). 



The normal is a straight line passing through a fixed point, 
namely the point of tangency, by § 225. 



i6o , ANALYTIC GEOMETRY 

Hence its equation must be of the form 
[i] y —y = s(yx' — x), by §57. 

in which x' andjj/' are the coordinates of the fixed point, here 
the point of tangenc}' ; ^ is the slope of the line, here the slope 
of the normal ; and x and y the coordinates of any point on the 
line, here the coordinates of any point on the normal PN. 

The normal is perpendicular to the tangent. by § 225. 

by § 62. 

by § 222. 



[2] 


Hence 


I -j- ^/ = 0. 


[3] 


But 


' ~a'y' 


[4] 


Hence 

cinrl 


, b'x' 

a' y 



Substituting this value of s into [i] , we get 

[6] y-y=-"^(x'-x). 

Now in [6] the x and y are the coordinates of any point on 
the normal PN. Hence [6] is the equation of the normal. 

by §39. 
Q. K.D. 

a^y 

227. Corollary. — The fraction — jr^} ^^ l^^ slope of the nor- 
mal. 

228. The Subnormal. — The subnormal is the distance 
measured along the transverse axis from the ordinate of the 
point of tangency to the normal. 

229. Corollary. — The length of the snb?iorfnal is — ^x\ 



a 



For the proof compare § 146. 



EXAMPLES 



I. Required the equation of the normal and the value of 
the subnormal in the following hyperbolas : 



9/— 4-^" 



X' 

a 



r 



THE HYPERBOLA i6i 

— 36, the point of tangency being (4, ord. +) 
I, the point of tangency being {-\/ a, o). 



PROPOSITION IX 



230. The tangent to an hyperbola bisects the interior^ and the 
normal the exterior angle between the focal radii of the poijit of 
tangency. 




I^et PM be the tangent and PN the normal to the hyper- 
bola at the point P. 

Let FP and F'P be the focal radii at the point P. 

We are to prove that 

PM bisects Z. F'PF, 

and that 

PN bisects Z. FPO. 

Since the point M is on the tangent, its coordinates -r= GM 
a.ndy=^ o must satisfy the equation of the tangent, by § 40. 

Hence substituting these values for the x and y of that 
equation, we get 

. bV b' x' 



[I] 



y = -irj ^-7CM. 

a~y a' y 



I62 




ANALYTIC GEOMETRY 






Hence 


ay- 


-d'x" = — CMb'x'. 






But 


ay - 


— b'x = — a 


'd\ 


by § 221, [2]. 


[2] 


Then 




a'd' = CMb'x', 




[3] 


or 




CM = "■; . 

X 






[4] 


Now 




MF = CF- 


-CM, 






or 




M^ = ae- 


a' 
x'' 


by§2i8, [2]. 


[5] 


hence 




MF=;(, 


ex' — a) . 




'6 


Also 




MF' = CF' 


+ CM, 






or 




MF' =ae-\- 


a' 
1 • 





X 

[7] UF'=^{ex'+a) 

[8] Then from [5] and [7] ^rv^,= ^^ 



MF' ex' + a' 

MF r 
[9] Hence MF'^T^' by § 220, [12] and [13]. 

[10] or MF : MF' ::V: /. 

Take PO = PF = r, 
and draw OF. 

Substituting this value of PF or r into [10], we get 
[11] MF : MF' :: OP : r. 

Hence OF II PM, by Geom. 24. 

[12] and ^ F'PM = ^ POF, by Geom. 8. 

[13] and Z^FPM =zlPFO. by Geom. 7. 

[14] But z:PFO =:Z:P0F. by Geom. i6. 

[15] Hence z^ F'PM=^ ^ FPM. 

Q. E. D. 
Again 

[16] z:F'PR + ^F'PM = ^FPM + ^FPN, by §225: 

[17] and z:F'PM=z1FPM. by [15]. 



THE HYPERBOLA 



163 



[18] Hence 
[19] But 
[20] Hence 

231. Corollary.- 
Compare § 149. 



^F'PR =z:OPN. 
^FPN =^OPN. 



by Geom. 4. 
Q. E. D. 



To draw a tangent to an hyperbola. 



232. A Chord. — A chord oi an hyperbola is a straight line 
terminated both ways by the hyperbola. 

233. The Bisector of a Complete System of Parallel 
Chords, — The bisector of a complete system of parallel chords is 
the line which contains all the middle points of those chords. 

234. The Diameter. — A diafueter of an hyperbola is that 
part of the bisector of a complete system of parallel chords 
which is bounded both ways by the hyperbola. 



PROPOSITION X 
235. The equation of a diafueter of an hyperbola is 

y= y-^cot(pjx, 

in which cp is the inclination of the system of chords bisected by 
diameter^ and a and b are the semi-axes. 




Let RS represent any one of a complete system of parallel 



l64 anal YTIC GEO me TR Y 

chords, qy its inclination, and P its middle point. Let IvN 
be the diameter which, when produced, bisects the system of 
chords represented by RS. 

We are to prove that the equation of LN is 

Let ;i: = CK and 7 ^ PK, 
^'=CH " y=RH, 
and r= PR = PS. 

[i] z^ RPO = z: PMK. by Geom. 8. 

[2] cos RPO = cos PMK = cos cp. 

[3] PO = r.cos RPO. by Trig. 2. 

[4] PO = r.cos q). 

[5] .;»;' = CK+PO = .r + r.cos cp. 

[6] RO = r.sin RPO. by Trig. i. 

[7] RO = r.sin q), 

[8] y = PK+ RO =jj/+r.sin ^. 

Now the point R is on the hyperbola, and hence its coordi- 
nates x^ andy must satisfy the equation of the hyperbola. 

by § 40. 
Substituting x' and j/' for the x and jk of that equation, 
we get 

[9] a^y'^ — d^x'- =^ — a'd^. by § 204. 

Substituting the values of x' and y found in [5] and [8] 
into [9], we get 

[10] ^^(jK+ r.sin cpY — d'{x -\- r.cos (pY ^^ — a'^b'^. 

Squaring the binomials and factoring with respect to r^ and 
r, we get 

[11] (a^sin^ cp — d'cos^ cp) r' -{- 2 (ay sin (p — d^x cos<^)r = 

Now since P is the middle point of RS, the two values of 
r in [11] must be equal to each other, and hence by the the- 
ory of quadratic equations 



THE HYPERBOLA 165 

[12] • 2(«^jj/ sin ^ — U^x 0.0^ cp^ ^^o. 

[13] Hence «^jf sin ^ = <^^jtrcos ^. 

b~ COS cp 

141 Hence j/ = — -^ x. 

■^ -^ a sm q) 

[15] Hence jf = ( — y cot <^j;t:. by Trig. 6 and 9. 

Now the X and jk of [15] stand for the coordinates of the 
point P. But, since RS represents any one of the system of 
parallel chords, P may be any point on their bisector PN, and 
since the diameter I^N is a part of the bisector, P may be any 
point on that diameter. 

Hence the x 2iX\Ay of [15] stand for the coordinates of any 
point on the diameter LN, which bisects the system of chords 
represented by RS, and therefore [15] is the equation of that 
diameter. by § 39. 

Q. E. D. 

236. Corollary i .^The diameter of an hyperbola is a straight 
line passing through the center. 

For proof compare § 154. 

237. Corollary 2. — If 6 be the inclination of any diameter^ 
and cp the inclifiation of its system of bisected chords, then 

b' 



tan 6 tan cp = 
For proof compare § 155. 



a^ 



i66 



ANAL YTIC GEO ME TR Y 



PROPOSITION XI 



238. If any diameter bisect a system of chords which are paral- 
lel to a second diameter, then that second diameter will bisect a 
system which are parallel to the first diameter. 




Fig. 78 

Let MN bisect a system of chords represented by RS, and 
let RS be parallel to OP. Let OP bisect a system of parallel 
chords represented by TU. 

We are to prove that 

TU II MN. 
Let 6 be the inclination of MN, 







9 


ii tt (i K^ 






6' 


" " " '' OP, 




and 


cp' 


( ( ( ( ( i ( ( 'T^TT 


[l] 


Then 




tan 6 tan <p=z ^^ by § 237. 

CI' 


[2] 


and 




tan 0' tan ^' := -^. by § 237. 

Ch 


[3! 


Hence 




tan Q tan q) =z tan 0' tan q)\ 
But RS 11 OP by Hypoth. 


[4] 


Hence 




^'=^, 


[5] 


and 




tan 6^ = tan (p. 



THE HYPERBOLA 



167 



Substituting this value of tan q) into [3], we get 
[6] tan Q = tan cp', 

[7] and = cp'. 

Hence TU || MN. by Geom. 9. 

and the system of chords parallel to TU will be parallel to MN. 

by Geom. 10. 
Therefore OP bisects a system of chords which are parallel 
to MN. 

Q. E. D. 

239. Conjugate Diameters. — Two diameters are said to be 
conjugate to each other when each bisects a system of chords 
which are parallel to the other. 

240. Corollary. — If Q be the inclination of any diameter, and 
6' the i?iclination of its conjugate, then 

b' 



tan 6 tan 6' = 



a' 



For proof compare § 158. 



PROPOSITION XII 



241. The tangent to an hyperbola at the extremity of any 
diam,eter is parallel to the conjugate of that diameter. 




Fig". 79 



Let OP be any diameter and MN its conjugate. Let PT 
be a tangent at the extremity of OP. 



i68 ANAL YTIC GEOMETR Y 

We are to prove that 

PT II MN. 

Let RS be any one of the system of chords bisected by OP. 

Let Q be the inclination of OP, 

6' " " '' " MN, 

cp " " " " RS, 

and ^' *' " " " PT. 

Let x' = CK and / = PK. 

We have RS || MN. by § 239. 

[i] Hence cp = B\ by Geom. 8. 

[2] and tan q) = tan d\ 

Since the point P is on the diameter OP, its coordinates x^ 
andj^' must satisfy the equation of that diameter which is 

[3] J=(-^cot^jx. by §235. 



Substituting x' 


and J/' for the x and y of 


this 


equation, we 


get 

[4] 




y=(^, cov<^x'. 






[5] 


Hence 








[6] 


But 


C^C\^ fTt 




by Trig. 9. 


cut tf ^ 

tan cp 


[7] 


Hence 


I _ a'y' 






tan cp b'^ x'' 




[s; 


and 


b'x' 

tan^= 2 f 

' ay 






[9l 


b^ x^ 
Hence by [2! tanO'=-^ — ;. 

or y 






]io] 


But 


, b'x' 
tan cp' — 2 r 

^ a' y' 




by § 222. 


[II] 


Hence 


tan 6' = tan cp\ 
Therefore PT ||MN. 


by 


[9] and [10]. 
by Geom. 9. 



Q. K. D. 



THE HYPERBOLA 



169 



242. Corollary i . — The two tangents at the extremities of any 
diameter are parallel to each other. 

243 . Corolla ry 2. — The four ta nge7i ts at th e extrem ities of any 
pair of conjugate diameters form a parallelogram inscribed 
within the two conjugate hyperbolas. 



PROPOSITION XIII 



244. Given the coordinates of the extremity of ajiy diameter to 
find the coordinates of the extremities of its conjugate. 




Let PQ and RS be any two conjugate diameters. Let OK 
and PK be given. 

We are to find OH and RH, OL and LS. 

Let x' = OK and / = PK, 
andjr'^= OH " jf" = RH. 

Draw the tangent PT. 

Since RS is a straight line passing through the origin, its 
equation must be of the form 

[i] , y— sx. by § 55. 

RS II PT. by §241. 

[2] Hence Z. ROX = Z PTX, by Geom. 8. 

[3] and tan ROX = tan PTX. 



I70 ANALYTIC GEOMETRY 

[4] But by [i] tan ROX — s, by § 53. 

[5] and tan PTX = ^,. by § 222. 

[61 Hence s = — , — -.. 

^ -^ a' y 

If we substitute this value of s into [i], the equation of RS 
becomes 

[7] y=^—i^- 

ay 

Now, since the point R is on the diameter RS, its coordi- 
nates x" and jk" must satisfy the equation of RS. by § 40. 

Hence substituting x" andy'^ for the x andj'of [7], we get 

[8] y" = -A,^"' 

Since R is also on the conjugate hyperbola RBB'S, its co- 
ordinates x" nndy^ must satisfy the equation of that hyper- 
bola, which is 

[9] ay — b'x' = a'b\ by § 2 1 1 , 

Hence substituting ^" andj^" for the x and y of that equa- 
tion, we get 

[10] «>"' — ^V'^=a^^^ 

Now, since in both [8] and [10] x" stands for OH and y" 
stands for RH, these equations are simultaneous and there- 
fore can be solved by algebra. 

Squaring both members of [8] , we get 

b^x'^ 

[11] y =-1-7-2-^ • 

ay 
Substituting this value of y^"^ into [10] , we get 

[12] a'^^.x''' — b'x"''=a'b\ 

ay 

[13] '~x"'^-x"'- = a\ 

"- ^'-' ay 



A^.ri 



[14] b'^x'^'x'" — ay'x'" = ay 

[15] and {ay' — b'x")x"' = — ay\ 



THE HYPERBOLA 171 

Now, since the point P is on the hyperbola PAA'Q, its co- 
ordinates x' and y must satisfy the equation of that hyper- 
bola, by § 40. 

Hence substituting x' and / for the x and y of that equa- 
tion, we get 

[16] ay — b''x'''=^ — arb\ by § 204. 

Substituting the value of the left hand member of this 
equation into [15], w^e get 

[17] «'^v = «y^ 

[18] Hence x" = ±i -^ y' =^OYi or OL. 

Substituting -| — j-y* for the x" in [8] , we get 
r- n ,, b~ x^ a , b , ^^^ 

[19] y"= -J—, -ry' = ^x' = RH. 

a y b a 

Substituting —y' for x*^ in [8] , we get 

[20] y' = _A^',= SI.. 

a 

245. Corollary. — All diameters are bisected by the center. 
For proof compare § 163. 



172 



ANALYTIC GEOMETRY 



PROPOSITION XIV 

246. The difference of the squares of any two conjugate 
diameters of an hyperbola is equal to the difference of the squares 
of its axes. 




Fig. 8r 

Let PQ and RS be any two conjugate diameters, and AA' 
and BB' the axes. 

We are to prove that 



PQ — RS = AA' — BB'. 

Let x' = CK andy = PK, 
^"=CH " y = RH, 
« = CA " <5 = CB, 
and«' = CP " ^' = CR. 



'l[ 




«'^ = ^'^+y^ 


by Geom. 26. 


'2 


and 


b'^^x'''^y"\ 


by Geom. 26. 


'.i. 


But 


tl2 /2 


by § 244, 


'a. 


and 


b^ 

y = a' * • 


by § 244. 



[5] Hence by [2] b" = ^y^ + ^ x'\ 



THE HYPERBOLA 173 

Subtracting the members of [5] from the corresponding 
members of [i] , we get 

[6] a" — b" .= ^''+y' — ^y — ^ x'\ 

[7] =.\.-^)^y\.-^). 

[8] =^.'^ + ^>. 

[9] =^^^ — ^^''• 

[10] =.^.^_,»)(^'_^y 

[11] «'^ — ^'^= («' — ^^)- 



!;2^/2 ^2^,/2 



,/? /,'2 _ /'^S A2^ "-^ ^^ 



Since the point P is on the hyperbola, its coordinates x' and 
y' must satisfy the equation of the hyperbola. by § 40. 

Substituting x' andy for the x andjj/ of that equation, we 
get 

[12] a'y'' — b''x''= — d'l)\ 

[13] b^x^' — a'y^a'b'. 

Substituting the value of the left hand member of this equa- 
tion into [11], we get 

[14] a'^ — b'^ = a' — b\ 

2 2 2 2 

[15] Hence PQ — RS = AA' — BB^. by § 245. 

Q. E.D. 



174 



ANALYTIC GEOMETRY 



PROPOSITION XV 



247. The parallelogram formed by tangents to tivo conJ2igate 
hyperbolas at the exti^emities of any pair of conjugate diameters 
is equal to the rectangle whose sides are equal to the axes of the 
hyperbolas. 




Fig. 82 

Let PQ and RS be any pair of conjugate diameters, and 
LMNO the parallelogram formed by tangents to the hyper- 
bolas at the extremities of these diameters. 

Let DEFG be the rectangle whose sides are equal to the 
axes AA' and BB'. 

We are to prove that 

LMNO — DEFG. 

Let « = CA and b = CB, 

y = CH " y = PH, 

;^;" = CK '' y = RK, 

Q^Z. PCH and ^ = ^ RCK. 

[i] LMNO = LO X NO sin LON. by Trig. 15. 

[2] Hence LMNO = \a'V sin RCP, by § 245. 

[3] or LMNO = \a'b' sin {cp — O). 



[4 
[5 

[6 

[7 

[8 

[9 

[lO 

[II 



THE HYPERBOLA 175 

sin \(p — 9) ^ sin cp cos Q — cos 9) sin 6. by Trig. 13. 



RK _ y" 
CR ~ b' 

CH_ ^ 
CP "■ a' ■ 
CK _ ^ 
CR "" ^' 

sm0 = ^ = ^. 



^'"^= CR=T 



cose = — =^, 



^°^^=CR= b' 



Hence sm (9^ - 0) = ^ ^ - ^ ^ =- -r^^. 

But ^" = -t-JK', by § 244. 

and /' = — x' . by § 244. 

a 



Substituting these values of ^" andj^" into [9], we get 

[12] sm(^-^).^ ^^, =___^____. 

Now since the point P is on the hyperbola PAA'Q, its co- 
ordinates jr' and J/' must satisfy the equation of that' hyperbola. 

by § 40. 

Substituting x^ andj^' for the x and y in that equation, we 
get 

[13] ay^ — ^*^'^ = — a-y^. by § 202. 

[14] Hence b'x"" —ay = a'd\ 
Substituting a^d^ for d'^x'^ — a^'' in [12], we get 

[15] ^>"('^-^) = «7^=7^- 

Substituting this value of sin {cp — 6) into [3], we get 

[16] IvMNO = 4a'd' 4rr = 4^^- 

a 

[17] But 4^3= 2«2^= AA'X BB', by § 245. 

[18] and AA'=GF. by Geom. 17. 



176 ANALYTIC GEOMETRY 

[19] Hence 4«/^=:GF X BB' = DEFG. 
Therefore from [16] and [19], we get 
[20] LMNO = DEFG. 



by Geom. 28. 



Q. E.D. 



The Directrix 

248. The Directrix. — The directrix of an hyperbola is a 
straight line drawn perpendicular to the X axis on the oppo- 
site side of the vertex from the focus, and at such a distance 
from the vertex that the distance from the focus to the vertex 
divided by the distance from the vertex to the perpendicular 
is equal to the eccentricity of the hyperbola. 



X- 





Y^ 



Fie:. 83 



FA 



In Fig. 83, if -— — = e, then DD' is the directrix. 
AF 

249. The Focal Distance. — The distance from any point 
on an hyperbola to the focus is called Va^ focal dista?ice of that 
point. 

FP is the focal distance of the point P. ' 

250. The Directral Distance. — The distance from any 



THE HYPERBOLA 



177 



point on an h3^perbola to the directrix is called the diredral 
distance of that point. 

PD is the directral distance of the point P. 



PROPOSITION XVI 

251. The ratio between the focal and directral distances of aiiy 
point on an hyperbola is constant and is equal to the eccentricity 
of the hyperbola. 




Let P be any point on the hyperbola, and let DR be the 
directrix. 

Join P to F and draw PM _L DR. 

We are to prove that 

PF 



PM 



e. 



[2] 



Draw the ordinate PK. 
Let X = CK, y = PK, ^ = HK, 
^ = C A, ^ = CF, and / = FH. 

/> = FH= FA + AH. 
FA 



AH 



e. 



by § 248. 



178 ANAL YTIC GEO ME TR Y 

[3] Hence AH = — — . by § 218, [2]. 

[4] Hence by [i] 

_^ . , ae — a . ae — a , ae — a 

p^YK-\ = <: — a -\ ^^ ae — a -\ = 

e e e 

ae^ — ae-\- ae — a 

e 

ae^ — a ai^e"^ — i) 



[5] Hence p 



[6] ;r=CH+HK = r— />+y = r+ {x' — p) . 

The equation of the hyperbola is 

[7] «y — h^x^ = — a'^b'. by § 204. 

Since the point P is on the hyperbola, the x and y of [7] 
may stand for the coordinates of that point. 

Substituting the value of ;i: given in [6] into [7], we get 

[8] ay-b\c^{x^-p)Y:=.-a^b\ 

[9] Hence a'y~ — b'^c' ^ 2c{x' — p) + iyx'—pY^^ —a'b\ 

b'^c^ 2<^V b'^ 

[10] f-^-^{x^-P)-^{^'-py^-b\ 

[11] But -^ =^^-i, by §218, [5]. 

[12] and c^z^a^e' by § 218, [2]. 

[13] Hence 

y — /^v — 2^(;p' —/)—(<?'— i)(x'— />)'=— ^^ 
[14] y— (^v— 2^(y— /)— ^H^'— /)^+(^-'— /)^=— ^^ 

[15] f-^^{x'-~py=b^e'^^2^{x^-p)^e\x'-py-b\ 

From [11] we get 

[16] b'^a\e^—l^. 

Substituting this value of b' and the value of p given in 
[5] into [15], we get 









THE 


HYPERBOLA 


[17] 




y 


+ (^' 


-py = rx'\ 


[i8] 


or 




PK + FK = ^-HK. 


J9. 


Hence 






PF = ^'HK, 


"20' 


and 






PF = ^HK = 


'21] 


Hence 






PF 

PA/r ^' 



179 



by Geom. 26. 



Q. E. D. 

252. Corollary i. — In an hyperbola the focal distance is greater 
than the directral distance. 

For e is greater than i. by § 219. 

Hence in § 251, [21], PF must be greater than PM. 

253. Corollary 2. — The distance from the center of an hyper- 
bola to the directrix is equal to — . 

e 

For in Fig. 84 

[i] Q,-^~c—ae, by §218, [2]. 

2 
[2] and FH = ^^— -^ by §251, [5]. 



e 
[3] Hence CH = CF — FH = ae 



ae^ — a a 



e 



i8o 



ANALYTIC GEOMETRY 



PROPOSITION XVII 

t 

254. The equation of the hyperbola when any pai?' of conjugate 
diameters are taken as the axes is 

hi which a^ and V are the semi-conjugate diameters. 




Let P be any point on the hyperbola. 

Let LM and NO be two conjugate diameters. 

Let LM be the new axis of abscissas, and NO the new axis 
of ordinates. 

Draw PS |i YY' and PK || NO. 

Let X = CS and j = PS. 
-r' = CK " ;/' = PK, 
«'^CM " ^' = CN. 
= Z^KCX, 0' = NCX. 

We are to prove that 

is the equation of the hyperbola referred to the diameters LM 
and NO. 



THE HYPERBOLA i8i 

[i] PR= PKsinPKR = y sin <9'. by Trig. i. 

and Geom. 1 1. 
[2] RS = KH = CK.sin 8 — x' sin B. by Trig, i . 

[3] Hence j = PR + RS =y sin 0' +x' sin 0. 

[4] Again RK= PK.cos PKR=:y cos Q\ by Trig. 2. 

and Geom. 11. 
[5] and CH = CK.cos = jt:' cos 0. by Trig. 2. 

[6] Hence jr = CH + RK = ;r' cos ^+y cos ^'. 

When XX' and YY' are taken as the axes, the equation of 
the hyperbola is 

[7] aY — b''x^^ — a'b\ by §204. 

Substituting for the x and y of this equation their values 
given in [3] and [6] , we get 

[8] a' ly sin^ 6' + 2xy sin 6 sin 6' + x" sin' O] 

— b'[y" cos' 6' + 2xy cos ^ cos 6'+ x" cos' ei~—a'b\ 
[9] Or (a' sin' ^—<5' cos' 6^)^" 

+ (a'sin'^' — ^'cos'^Oy 
+ 2(a' sin 6 sin 6*' — <^' cos cos 0')xy =1 — a^b'\ 

b^ 
[10] But tan tan ^' = — ^ . by § 240. 



["] 


Hence 


2 . ZD 


"^ ^^''^-tan^^" 


[12] 


or 


, sin r 2 cos 0' -u ^ • ^ 

a~ ^ = ^'-^ — ^,. by Trig. 6. 

cos u sm fc/' ^ 


[13] 


Hence 


^' sin Q sin 6' = ^' cos 6^ cos ^', 


!i4; 


and 


a^ sin ^ sin 0' — ^' cos ^ cos Q' = 0. 


[15] 


Substituting for this binomial in [9], we get 


[16] 






(<2' sin' 


6 b'cos' 0)x"+ (^'sin' 6' ^'cos' 6')y'=: a'b\ 



which is the equation of the hyperbola when LM and NO are 
taken as the axes of coordinates. 

Since the point M is on the hyperbola, its coordinates a' and 
o must satisfy the equation of the hyperbola. by § 40. 



i82 ANALYTIC GEOMETRY 

Substituting these values for the x^ and_y' of [i6], we get 
[17] {a" sin^ e — b' QOse)a'"- — — a'b\ 

a-b' 



[18] }Ience a^ sin^ — b" cos^ 



a'^ 



Now, if instead of taking the point P, we had taken any point 
P' on the conjugate hyperbola, and had taken YY'for the rectan- 
gular X axis and XX' for the rectangular Y axis ; and had 
taken NO for the oblique X axis and MI^ for the oblique Y 
axis ; and also had taken BCN for d and BCM for Q' , then in 
the same wa}^ that we obtained [16] we would have obtained 

[19] (CB sin' BCN — CA cos" BCN)CK' + 



(CB sin' BCM — CA cos' BCM) P'K' = — ^^^^^ 

If we take Mly for the X axis and NO for the Y axis, then 
P'K' = CT may be represented by x" and CK'=: P'T by y'\ 
and [19] wnll become 

[20] 
(^'cos'^' — «'sin'0')y"+(^'cos'^ — «'sin'a):r"'z=— «'<5'. 

In [20] x" and J/" stands for the coordinates of any point 
on the conjugate hyperbola when LM is taken as the X axis 
and NO is taken as the Y axis. 

Since the point N is on the conjugate hyperbola, its coor- 
dinates o and b' must satisfy [20]. 

Substituting these values for the x" and jv" of [20] , we get 

[21] (<^' cos-^ & — a' sin' e')b" — —a'b\ 

[22] a' sin' 0'— /^'cos' Q' = ^. 

Substituting the right hand members of [18] and [22] into 
[16], we get 

2 L-2 2 7,2 

[23] — -^ ^" + -j^y — — ^'^'• 

[24] Hence a'y — b^'x'' =^ — a^'b'\ 

In [24] the x' and y' stand for the coordinates of any point 



THE HYPERBOLA 



183 



on the h3^perbola PML, and hence [24] is the equation of that 
hyperbola. 

Now since the oblique axes only are to be used, there is no 
further need of the accents over the x and jj/, and hence the 
equation of the hyperbola may be written 

[25] 



^'y _^/V = — «'^^'^ 



Q. E.D, 



PROPOSITION XVIII 

255. When a7iy pair of conjugate diameters are taken as the 
axes of coordinates^ the equation of the tangent to an hyperbola is 

in which x* and y are the coordinates of the point of tangency, 
and a' and b' are the semi-conjugate diatneters. 




IrCt AA' and BB' be any two conjugate diameters. 

Let AA' be the X axis and YY' the Y axis. 

Let PT be tangent to the hyperbola PAA'D at the point P 

Let x' = CK and / = PK, 
a' = CA '' b' = CB. 

We are to prove that 



y — y 

is the equation of PT. 



b"x' ,_ 



i84 ANALYTIC GEOMETRY 

Let PM be a secant cutting the hyperbola at P and P'. 

Let y = CH and /' = P'H. 

Since the secant PM is a straight line passing through two 
fixed points, its equation must be of the form 

Since P is on the hyperbola, its coordinates x' and y' must 
satisfy the equation of the h3^perbola when A A' and BB' are 
taken as the axes. 



Substituting x^ and j/' for the x and y of that equation, we 



cet 



[2] ajy — ^'V- = — a'''U\ by § 254. 

Similarl}^ since the point P' is also on the hyperbola, we 
may also get 

[3] ^.y — b'^x"^ = — a"b'\ 

Now proceeding as in § 134, we get 



Wx 

[4] y— J^=^7y <-^'— ■^)- 



Q. K. D. 



255^, Corollary, — The equation of the tangent may also be 
written 

[5] «'!r> —b"x'x = — a''b'\ 



PROPOSITION XIX 

256. When any pair of conjugate diameters are taken as the 

axes^ the equation of the chord which joins the points of tangeiicy 

of two tangents drawn to an hyperbola from the same point 

without it is 

ayy' — b"xx' = —a"b'\ 

in which x' and y' are the coordinates of the point from which 
the two tangents are drawn, and a' and U are the semi-conjtigate 
diameters. 



THE HYPERBOLA 185 




Fig-. 87 

Let PT and P'T be two tangents drawn to an hyperbola 
from the same point T. 

Let PP' be the chord joining the points of tangenc3\ 

Let any two conjugate diameters LM and NO be taken as 
the axes. 

Let x' = CR andy = TR, 

^' = CM " /^' = CN. 

We are to prove that 

is the equation of the chord PP'. 

Let x^' = CK and y" = PK. 
x'^'^CK " y"=P'H. 

The equation of PT is 

[i] y —y=^^ny7(^" — ^)- by §255. 

[2] Hence a'yy — a'y'" = d"xx'' — 3' V'^ 
[3] and a'yy" — d"xx" = a'y" — d"x"\ 

Since the point P is on the hyperbola, its coordinates x'' 
and jv" must satisfy the equation of the hyperbola. 

Hence substituting x" and y" for the x and y of the equa- 
tion of the hyperbola, we get 



i86 ANALYTIC GEOMETRY 

[4] «'y — ^' V = — a'^b'\ 

Now substituting — a''^b'^^ for the right hand member of [3] , 
we get 

[5] «'vy'— ^"•^•^" = — ^"^", 

for the equation of PT. 

Similarly we may show that the equation of P'T is 

[6] a"yy"—b"xx''' =—a''b'\ 

Since the point T is on the tangent PT, its coordinates x' 
andy must satisfy [5], the equation of PT. 

Hence substituting x' andj/' for the x andjK of [5], we get 

[7] «'>V' — ^"•^'^" = — a^'b'\ 

Similarly since the point T is also on the tangent P'T, we 
get 

[8] a'»"' — ^'V;r"' == — a'''y\ 

Now equation 
[9] a'yy — b'\x'x——a'''b''' 

is the equation of a straight line. by § 67. 

But the coordinates x" 2ind.y" of the point P will satisfy this 
equation, for if they are substituted for the x and y in it we 
get [7]. 

Hence the straight line represented by [9] must pass 
through the point P. by § 41. 

The coordinates x'" and j/'" of the point P' will also satisfy 
[9] , for if they are substituted for the x and^j/ in it we get [8] . 

Hence the straight line represented by [9] must also pass 
through the point P'. by § 41. 

Hence, since the straight line represented by [9] passes 
through both the points P and P', it must be the chord PP'. 

Therefore [9] must be the equation of the chord PP'. 

Q. K- D. 

257. When the transverse axis of the hyperbola is taken as the 
X axis and the conjugate axis as the Y axis, the equation of the 
chord becomes 

a^yy' — b^xx' == — a^b' . 



THE HYPERBOLA 



187 



PROPOSITION XX 

258. The two tangents at the extremities of any chord of an 
hyperbola meet on the diameter which bisects that chord. 




Fig. 87a 

Let PT and P'T' be tangents to the hyperbola at the ex- 
tremities of the chord PP'. 

Let LM be the diameter which bisects that chord. 

We are to prove that that PT and P'T' meet on the diame- 
ter LM. 

Let ON be the diameter which is conjugate to LM. 

Let LM be taken as the X axis and NO as the Y axis. 

Let a' = CM, b' = CN, and x' = CK. 

If R be the point where PT cuts LM, the X axis, then 
as in [3] of § 230, we get 



[I] 



CR = 



a' 



X' 



by § 45. 



Similarly, if R' be the point where P'T' cuts LM, the 
X axis, then 

[2] 



[3] Hence 



CR' = 
CR:= 



X 

CR', 



1 88 



ANALYTIC GEOMETRY 



which shows that both tangents meet the diameter LM at the 
same point. 

Q. E. D. 

PROPOSITION XXI 

259. If two tangents be drawn through the extremities of any 
focal chord of an hyperbola, 

(7) the two tangeiits will meet o?i the directrix ; 

{2) the I i7ie joining the intersection of, the two tangents to the 
focus will be perpendicular to the focal chord. 




Fig-. 88 

Let PT and P^T' be two tangents drawn to the hyperbola 
at the extremities of the focal chord PP'. Let R be the inter- 
section of the two tangents. 

Let DD' be the directrix. 

We are to prove that R will be on the directrix DD'. 

Let -r'= CK andy = RK, 
i? = CA '^ b =CB. 

Let X and y be the coordinates of any point on the chord PP'. 
The equation of PP' is 
[i] a'^yy' — b^xx' z=z — a'^b'^, by §257. 

in which x^ and y' are the coordinates of the point R. 



THE HYPERBOLA 189 

Since the point F is on the chord PP', its coordinates x:=^ ae 
and jK = o, must satisfy the equation of that chord. by § 40. 
Substituting these values for the x 2.nd,y of [i], we get 
[2] — b'aex' ■=. — a^F' . 

[3] Hence x' = — . 

But since by § 253, — is the distance from the center to 

e 

the directrix, R the intersection of the two tangents, must be 

on the directrix. 

Q. E. D. 

Again, since RF is a straight line passing through the two 
fixed points R and F, its equation must be of the form 

[4] y__y=-^J;zZ'(y_^). by §58. 

In [4] let Jtr" and /' stand for the coordinates of the point 
F, and x' and j/' stand for the coordinates of the point R. 

Then x*^ = ae and jj/" = o. 

Substituting these values of x'' andjF" into [4], we get 

[5] y-^=<^(-'— )• 

But jt' = — . by [3]. 

Substituting this value of x' into [5], we get 

[6] V~y~ ^^ ^ {x'—x), 

^ -* . "^ a — ae 

[7] Hence y' — J^ = -27 ' — 2t(^-^' — •^). 

L/J ^ ^ ^2^j — ^Z^ 

which is the equation of the line RF. 



[8] By [3] 



a 



e 



Substituting this value of x' into [i] , we get 
[9] ayy^b' -^x — a^b\ 



I90 ANALYTIC GEOMETRY 

\_ii\ Hence a^ey^y = U^ax — a^^V, 

Pi 21 and r = r ix — ae). 

^ -" -^ aey 

[13] But b' — a\e' — Y):=^ — a\i — e''). by §218, [5]. 

r T XT d'^i—e") 

14 Hence r = — — ^, — {x — ae). , 

aey 

which is the equation of the chord PP'. 

Let s = the slope of the line RF, 

and/= '' '^ " " chord PP'. 

From [7] we get 

[■5] '= a\7-ey ^''^^^- 

From [14] we^get 

[16} .'= ^^^^. by §53. 

[17] Hence ^^' = , , — ^r X ^^ 1 = — i, 

*- -^ a {1 — e) aey 

[18] or I -[- 5/ = o. 

Therefore RF and PP' are perpendicular to each other. 

by § 62. 
g, K. D. 



PROPOSITION XXII 

260. The locus of the intersectio7i of two tangents to an hyper- 
bola which are perpeiidicular to each other ^ is a circle whose center 
is at the origin, 

Let PT and P'T be two tangents to an hyperbola at the 
points P and P'. Let them be perpendicular to each other at 
the point T. 

Let x' = CK andy = PK. 

Let P and P' move along the hyperbola in such a wa}^ that 
Z^PTP' shall always be a right angle. 



THE HYPERBOLA 

Y 



191 




Fiff. 



We are to prove that the locus of the point T will be a circle 
whose center is at the origin. 



The equation of the tangent PT is 

to 



y —y — -1—/ {x' —x) . 

-ay 



by § 221, 



[2] Hence tz» — <?y' = ^ V;i: — ^ V^ 
[3] and ' ayy = b'x'x + «>'' — b''x'\ 

Since the point P is on the hyperbola, its coordinates jc' and 
y^ must satisfy the equation of the hyperbola, 

[4] Hence a'y'' — ^V= ^^ ~a'b\ by § 204, 

Substituting this \^alue of ay^- — b'^x'^ into [3], we get 



C5] 

[6] Hence 



a^yy = a^x^x — «^3' 



b'^x' 
^ ay' 



y 



[7] Now y = ^-r, = \b'-^,, which by 



[4] 



V' 



=\«5 



T2 1'> 9 / 



\ 



i4 2 

ox 



a'b'y 

~ay^ 



[8] 



(^^ / d'^x'^ I (b'^ x'Y 

Hence -7-= V^'^-T2 — ^' = \^ A^~/ J — '^^ 



192 ANALYTIC GEOMETRY 

Substituting this value of — y into [6] , we get 



•^ ay ^ \a- y J 

h^ x' 

Now let s^=^—y — i. 

« y 

Then [9] becomes 



[10] y =:l sx — ^«V — ^', 

which is the equation of any tangent PT to an hyperbola. 



[11] Let y = s'x — ^d's'' — b\ 

be the equation of the tangent P'T, which is perpendicular 
toPT. 

Then 

[12] i + 5/ = o. by § 62. 

[13] Hence j' = . 

Substituting this value of ^ into [n], we get 

[14] y = J — \-J-b\ 

for the equation of P'T. 

By transposition [10] and [14] become 



[15] y — sx^ — -j/^V — b^, 



[16] and yJr^=-\^-b\ 



\ a' 



By squaring [15] and [16] become 

[1 7] y — 2sxy + s^x"^ = « V — b^, 

[18] and y + 2^ + ^=^-^'. 

Clearing [18] of fractions, we get 

[19] s~y' -\- 2sxy -\- x^ = d^ — s'^b'^ 



THE HYPERBOLA 193 

Adding [17] and [19], we get 

[20] (I + /)y + (I +/)^^zz= «*xi + .o-(i + s')b\ 

[21] Hence y'' ^ x' ^ a" — b\ 

in which x and_>^ are the coordinates of the point T. by § 48. 
Hence [21] is the equation of the locus traced out by T. 

i by § 39. 

Now let a' — y — r\ 

[22] Then by [21] ;i:'+y = r^ 

which is the equation of a circle whose center is at the ori- 
gin, by § 113, 
Therefore the locus traced out by T is a circle whose center 
is at the origin. 

Q. E. D. 

PROPOSITION XXIII 

261. If any chord of an hyperbola pass through a fixed point 
and tangents be drawn at its extremities^ and if the chord be made 
to revolve about the fixed point as a pivot, then the locus of the in- 
tersection of the two tangents will be a straight line whose equa- 
tion is 

a''yy' — b^'xx'=—a^'b^\ 

in which jr' and y' are the coordiriates of the fixed point about 
which the chord revolves, and a' and b' of the semi-conjugate di- 
ameters which are taken as axes. 




Fig. 90 



194 ANALYTIC GEOMETRY 

Let PS be a chord passing through the fixed point R, and 
let PT and ST be tangents drawn at its extremities. 

Let LM and NO be two conjugate diameters taken as axes. 
Let x' = CH and / = RH, 

Let PS revolve about R as a pivot. 

We are to prove that the locus traced out by T is a straight 
line and that its equation is 

ayy — d"xx' = — a"d'\ 

Lety = CK and y' = TK. 
The equation of the chord PS is 
[i] a'yy — d''xx'' =z—a"d'\ by § 256. 

Since the point R is on this chord, its coordinates x' ar\dy 
must satisfy the equation of this chord. by § 40. 

Substituting x' and_r' for the x andj' of [i] , we get 

[2] ayy — U'^x'x" = — a'''b'\ 

Now as PS revolves about R, T will trace out a locus. 
Moreover [2] will be satisfied by the coordinates of the point 
T wherever it may be as it traces out this locus. 

Hence the x" and y of [2] stand for the coordinates of 
every point on the locus traced out by T. 

Therefore [2] must be the equation of that locus, by § 39. 

This locus must be a straight line. by § 67. 

But since T is any point on this straight line traced out b}^ 
the intersection of the tangents, we may drop the accent marks 
from its coordinates and write them x and y, hence [2] may 
be written 

[3] «'!ry — b'^xx' = — a''b'\ 

Therefore the locus traced out by T is a straight line and 
its equation is 

[4] a'^yy — b"xx' — — a"b'\ 



THE HYPERBOLA I95 

262. Supplemental Chords. — Two chords drawn from the 
same point on an hyperbola to the extremities of any diame- 
ter are called supplemental chords. 

PROPOSITION XXIV 

263. If a chord be parallel to any diameter of an hyperbola the 
supplemental chord will be parallel to the conjugate diameter. 




^E' 



Fig' 91 

Let MC and MC be two supplemental chords drawn to the 
extremities of the diameter CC. 

Ivet DD' and KB' be two conjugate diameters. 

Let MC II EH'. 
We are to prove MC || DD'. 



]l^ 




[2] 


But 


[3] 


Hence 


[4] 


and 



CO _ C K 
C'C ~C'M' 

C^O_, 

C'C ""'• 

CK_ , 
CM'"^' 



by Geom. 23. 
by § 245. 



C'K = iCM. 

Hence the diameter BE' || MC bisects a system of chords 
II MC. But by hypothesis the diameters BE' and DD' are 
conjugate ; and therefore BE' bisects ^ a system of chords 
which are || DD'. 

Therefore MC H DD'. by Geom. 10. 

Q. E. D, 



196 



ANALYTIC GEOMETRY 



Polar Equation of the Hyperbola 



PROPOSITION XXV 



264. When the right harid focus is taken as the pole, the polar 
equation of the hyperbola is 

_ aie^ — i) , 



I — e COS 



e 



in which a is the semi-transverse axis of the hyperbola, e its eccen- 
tricity , r the radius vector of any point on it, and d the vectorial 
angle of that point. 




Fig. 92 

Let F be the pole, XX' the initial line, and DD' the direc- 
trix. 

Let P be any point on the hyperbola. 

Let Q'^Z. PFX, r = FP, and e = the eccentricity. 

We are to prove that 

I — e cos d 
is the polar equation of the hyperbola. 

DrawPD_LDD' and PK J_ XX'. 
[i] PF == ^PD. by §251. 

[2] Hence PF = ^(EF+FK), by Geom. 17. 

[3] or PF = ^.EF+ ^.FK. 



THE HYPERBOLA 197 

[4] But EF = ^^^^, by §251, [5]. 

[5] and FK = r cos Q. by Trig. 2. 

Hence from [3] , [4] and [5] , we get 
[6] r ^ a{e^ — i ) + <?r cos Q . 

[7] Hence r(i — ^ cos 6^)=^^(^^ — i), 

[8] and r = -^ ^. 

*- I — e cos C7 

Q. E.D. 

265. Corollary. — When the left ha7id focus is taken as the 

pole, the polar equation of the hyperbola becomes 

a(e^ — I ) 

r :^= — • . 

e cos — I ' 

EXA.MPLES 

1. What is the equation of the hyperbola conjugate to 
2^y^ — i6:r^ = — 400 ? 

2. What is the eccentricity of the h3^perbola conjugate to 
i6y^ — gx^ = — : 144 ? 

5. Find the equations of the two tangents to 4y^ — i2x^ =^ 
— 48, at the upper and at the lower extremities of the param- 
eter. Ans. y =z 2X — 2. 

jK = — 2;r + 2 . 

4. How far is it from the intersection of these tangents to 
the directrix ? 

5. What is the angle between the two tangents drawn to 
i6y^ — gx^ = — 144 at the extremities of the parameter ? 

6. Find the equations of the two tangents drawn to 
^y — b^x^ = — a'^b^ at the extremities of the parameter. 

Ans. y =^ ex — a. 
y =■ — ex -\- a. 

7. Find the angle betweien these two tangents and the dis- 
tance of their intersection from the directrix. 

2e 

Ans. tan~' — -. 

I — ^ 

Distance = o. 



198 ANAL YTIC GEOMETR Y 

8. Where does the tangent whose inclination is 45° touch 
25 jj/" — i6x* = — 400 ? Ans. At x' = 8-^. 

y = i\- 

9. Where does the tangent whose inclination is 45° touch 
«V" — b'^x'^ = — a'b'^ ? 

Ans. At X = 



1/ «^ — ^■ 



V a' — b' 

10. Where does the tangent at the vertex of its conjugate 
cut i6jj^^ — 9jr- 1= — 144? Ans. :ir = db 5.6. 

11. Where does the tangent at the vertex of its conjugate 
cut a^y^ — b'x'^ = — a^b^ ? Ans x = a^ 2. 

12. Where does the tangent drawn from the upper vertex 
of its conjugate touch \6y~ — <^x^ = — 144 ? 

Ans. At ;r = 4^ 2. 

13. Where does the tangent drawn from the upper vertex 
of its conjugate touch ^y — b'^x' = — a'^b^. 

Ans. At X ^= a. 

jy = — b. 

14. What are the coordinates of the upper end of the param- 
eter of ay- — b^x'^ = — a"b^ ? 

15. The equation of the tangent drawn through the upper 
extremity of the parameter of an hyperbola is jf = ix — 4. 
What is the equation of the hyperbola ? 

Ans. i6j/^ — 9^^ = — 144. 

16. The tangent drawn through the upper end of the 
parameter of an hyperbola whose . semi-transverse axis is 5, 
passes through the vertex of the conjugate hyperbola. What 
is the equation of the first hyperbola and what is its eccen- 
tricity ? Ans. 2^^ — 25jr^ = — 625. 

^ = 1/ 2. 



THE HYPERBOLA 199 

17. From what point must two tangents to qjk^ — i6;r* ^ 
— 144 be drawn that the}^ may touch the hyperbola at the ex- 
tremities of a chord which passes through the vertex of the 
hyperbola and the upper end of the parameter ? 

Ans. From ^ := 3. 
_y = 2. 

18. iVbout what point must a chord of 36)/^ — i^x' = — 576 
revolve in order that the locus of the intersection of the two 
tangents drawn through the extremities of this chord may 
bisect the positive halves of the axes of the hyperbola ? 

Ans. x' = 12. 
/ = —8. 

19. About what point must a chord of ;^6y — i6x^ = — 576 
revolve in order that the locus of the intersection of the two 
tangents drawn through the extremities of this chord may 
bisect the positive half of the transverse axis and the nega- 
tive half of the conjugate axis ? Ans. x' z= 12. 

y = 8. 

20. About what point must a chord of <zy — d^x"=. — a^d^ re- 
volve that the locus of the intersection of the two tangents 
drawn through the extremities of the chord may bisect the 
positive half of the transverse axis and the negative half of 
the conjugate axis? Ans. x' = 2a. 

y = 2d. 

21. About what point must a chord of ay^ — d'^x^ =^ — a^d' 
revolve in order that the locus of the intersection of the two 
tangents drawn through the extremities of this chord ma}* 
bisect the positive halves of the axes of the hyperbola ? 

Ans. About x =z 2a. 
y = — 2d. 



CHAPTER XI 

Asymptotes 

266. The Asymptotes. — The diagonals of the rectangle 
constructed upon the axes of an h3^perbola are called the 
asymptotes of the hyperbola. 




In Fig. 93 LM and NO are the asymptotes. 



PROPOSITION XXVI 
267. The equations of the asymptotes of an hyperbola are 

b 
a 



and y 



a 



X, 



in which a and b are the semi-axes of the hyperbola. 

For in Fig. 93 the asymptote LM is a straight line passing 
through the origin. by § 266. 

Hence its equation must be of the form 
[i] y — sx. by § 55. 

[2] Hence s = tan MCA. by § 53. 







ASYMPTOTES 


201 


[3] 


But 


. ....A MA b 
tan MCA = 7— r- = — . 
C A a 


by Trig. 3. 


[4] 


Hence 


b 
a 





Substituting this value of ^ into [i], we get 

b 



[5] 



y 



X, 



a 



for the equation of the asymptote LM. 

Similarly it may be shown that the equation of the asymp- 
tote NO is 

[6] y= X. 

a 



PROPOSITION XXVII 

268. As the asymptote extends outward from the center of the 
hyperbola, the distance between it and the hyperbola continually 
approaches o, and if the asymptote be made long enough this dis- 
taiice may be made less than any assignable quantity, but the 
asymptote can never touch the hyperbola. 




Let lyM be one of the asymptotes of the hyperbola. 
Let R be any point on it and draw RK JL XX' and cutting 
the hyperbola at the point P. 

Through P draw PSJ_ LM. 

Let X = CK and y = PK. 



202 ANALYTIC GEOMETRY 

We are to prove that as the point S is taken further and fur- 
ther from the center of the hyperbola, PS continually ap- 
proaches o, can be made less than any assignable quantity, 
but can never become o. 

The equation of I^M is 

b 
[ij J = — X, by § 267. 

[2] or RK = ^ CK. 

The equation of the hyperbola is 
[3] «y ~ ^'-^^ = — ^'^'. by § 204. 

[4] Hence y^ — \/-^^ — ^'' 

b 



[5] or PK = ^^eK- 



a\ 



[6] Hence RP = RK — PK = — (CK — \'CK — «') • 
[7] Hence 

b CK — CK + «' ab 



RP = 



CK + \/cK — a' C K + ^CK— «^ 



[8] Now PS = RP.cos RPS, by Trig. 2. 

[9] and z:RPS = z::LCA. by Geom. 12. 

[10] Hence PS = RP cos I^CA. 

Substituting the value of RP given in [7] into [10], we get 

[11] PS=: "^ ^ cos LCA. 



CK + 



VcK — «.' 



Now let the point R move along the asymptote away from 
the center. 

The quantities ab and cos LCA do not change their values 
in consequence of this motion of the point R. The only part 



ASYMPTOTES 203 

of the right hand member of [11] that does change its value 
is the denominator of the fraction, which increases continually. 

Since the numerator of this fraction is constant, and its de- 
nominator is increasing, the value of the fraction is con- 
tinually approaching o. 

Since the denominator can be made as large as we please, 
the value of the fraction can be made as small as we please. 

Since the numerator is constant, the value of the fraction 
can never become o. 

Therefore, as the asymptote extends outward from the center 
the distance PS between it and the hyperbola continually ap- 
proaches o, can be made less than any assignable quantity, 
but can never become o. Q. E. D. 

269. Corollary. — As the asymptote extends outwards from the 
center^ the distance from, it to the conjugate hyperbola continually 
approaches o, but it can never touch the conjugate hyperbola. 

It is obvious that any other straight line passing through 
the center must meet either the hyperbola or its conjugate. 
Hence we may define the asymptote as follows : 

270. The Asymptote. — The asymptote of an hyperbola is 
a straight line passing through the center, whose distance 
from the hyperbola can be made less than any assignable 
quantity by taking a point on the hyperbola far enough 
from its center, but can never be made o. 



PROPOSITION XXVIII 

271. The tangent to an hyperbola can be made to coi7icide as 
nearly as we please with the asymptote by moving the point of 
tang ency far enough froTn the ceyiter. 

For, the equation of the tangent is 

b'' x^ 

[l] y^ —y= ^{x' — x). by §221. 

b'^x^x b^ 
[2! Hence y = — ^ — -. r . as in § 260, [61 . 



204 ANALYTIC GEOMETRY 

Since the point of tangency is on the h5^perbola, its coor- 
dinates x^ andjj/' must satisfy the equation of the h^^perbola. 

by § 40. 

[3] Hence ay' — b''x"= — d'b\ by § 204. 



[4] and -^' = ^\/^" 



a\ 



Substituting this value of jj/' into [2], we get 
r -, b- x^ }f 



a 



\ t2 5 

-\X — a 



[6] Hence jy= — -x 



2 



Now let the point {x, y) remain fixed and the point of tan- 
gency move away from the center continually. 

[7] Then limit r = limit ( — x — __-_—_- — — V 

V a I ^ y J 



x" 



The value of -* / j ^ will continually approach i, and 

^ x'^ 

the value of — ^ will approach o. 

/ a^ 
Thus as x' and y' increase indefinitely, the limit of a/ i ^ 

. . b' 
IS I , and the limit of — r =0. by Geom. 19. 

When these limits are substituted into [7], it becomes 
[8] y^-^x, 

which is the equation of the asymptote. 

Hence by moving the point of tangency far enough from the 
center we can make [6] coincide as nearly as we please with 
[8]. 

Q. E. D. 



ASYMPTOTES 



205 



PROPOSITION XXIX 

272. When any pair of conjugate diameters are taken as the 
axes of coordinates^ the equatio?is of the asymptotes are 

a' 



and y 



U 



a 



r-^» 



in which a^ and h' are the semi-conjugate axes. 




Fig:- 95 

Let PQ and RS be the asymptotes. 
Let LM and NO be any two conjugate diameters. 
Let LM be the X axis and NO the Y axis. 
Let a' = CL and b' = CO. 
We are to prove that 



[I] 

is the equation of PQ, 

[2] and 
is the equation of RS. 



y = -yx 
a 



d' 



X 



2o6 



ANALYTIC GEOMETRY 



When LM and NO are taken as the axes of coordinates, 
the equation of the hyperbola is 

a'y — ^'V = — «''3'^ by § 254. 

Since this equation is of exactly the same form as [3] , § 268, 
and since [i] is of exactly the same form as [i] of the same 
section, then by the same method as was used in that section 

b' 
we may prove that the distance from jk = — r ^ to the hyper- 
bola is continually approaching o, can be made less than any 
assignable quantity, but that it can never become o. 

Hence 

b' 

is the equation of the asymptote. 
Similarly it may be shown that 



by § 270. 



y 



N 



X 



a' 



is the equation of the other asymptote. 



Q. K. D. 



PROPOSITION XXX 

273. The asymptote is the diagonal of every parallelogram 
whose sides are tangents parallel to two conjugate diameters. 




Fig. 96 



ASYMPTOTES 207 

Let LM and NO be two conjugate diameters. 

Let LM be the X axis and NO the Y axis. 

• Let a' = CL and V = CO. 

Let PQ and RS be the asymptotes. 

Draw the tangents LP and OP parallel to the conjugate 
diameters LM and NO. 

We are to prove that the diagonal PC of the parallelo- 
gram PLCO coincides wnth PQ, and the diagonal KC of the 
parallelogram KOCM coincides with RS. 

Let /^z:PCL, 

and Go=^OCh. 

The equation of the diagonal PC will be 

sin / 



I 




sin {CO — /) 


by § 76. 


[2] 


But Ge9- 


-/=z:ocprrz:cPL. 


by Geom. 7. 


[3] 


Hence 


sin PCL 

V = X. 

^ sin CPL 




[4] 


PL ; 


: CL : : sin PCL : sin CPL. 


by Trig. 14. 


[5] 


Hence d' ; 


\ a' \: sin PCL : sin CPL, 





V sin PCL 
M °'- V = sha^PL- 

Substituting the left hand member of this equation into [3], 
we get 

[7] ^=^^- 

But this is the equation of PQ. by § 272. 

Therefore the diagonal PC coincides with the asymptote PQ. 

Similarl}^ it may be shown that the diagonal CK coincides 
with the asymptote RS. 

Q. E.D. 



208 



ANALYTIC GEOMETRY 



PROPOSITION XXXI 

274. When the asymptotes are taken as the axes of coordinates^ 
the equation of the hyperbola is 

a' + b' 
xy= 

in which a and b are the semi-axes of the hyperbola. 




Fig. 97 

Let IvT and UV be the asymptotes of the hyperbola. 
Let LT be taken as the Y axis and UV as the X axis. 
Let P be any point on the hyperbola, and draw PM _L AA' 
produced and PM' || LT. 

Letjr = OM and JK = PM, 
y = OM' " y = PM', 
« = 0A " /5 = 0B. 

We are to prove that 

a' + U' 



xy — 

•^ 4 




is the equation of the hyperbola. 




[i] Now tanROA = — , 
•- -J a 


by Trig. 3. 


b 

'2] and tan SOA = — . 
■-J a 


by Trig. 3. 



ASYMPTOTES 209 

[3] Hence tau ROA = tan SOA, 

[4] and ^ROA = z::SOA. 

Let /? ^ zl ROA = Z. SOA. 

[5] jf ==. ON + M'Q = OM'cos SOA + PM' cos PM'Q, 

by Trig. 2. 
[6] or jr = ;r' cos /? + y cos ^^ (:r' +_>/') cos ^. 

[7] Again j = PQ — M'N = 

PM'sinPM'Q— OM'sin SOA. 

by Trig. i. 
[8] or jK=ysin/? — .r'sin/?=(y — ^') sin/?. 

Now when the axes of the hyperbola are taken as the axes 
of coordinates, the equation of the hyperbola is 

[9] d'f — b''x'' = — a'b\ by §204. 

Substituting the values of x and y given in [6] and [8] 
into this equation, we get 

[10] a\y'—xy-,Wft — b\x'-\-y'yQO<^'P = —a''b\ 

[11] Hence (^- cos'/?— a^ sin^y^) (;r'^ +y^) + 
2 (^' cos' /? + «' sin' ^')x'y' = a'b\ 

[12] But tan^=-^ = — , by[i]. 

\n,\ or — — ^ = — . by Trio:. 6. 

■- ^-' cos ^ a ^ o 

[14] Hence a sin /3 = b cos /3. 

[15] and a' sin' /? = ^' cos' /?. 

Substituting these values into [11] , we get 

[16] 4a' x'y sin' /3 =: a'b\ 

[17] 4xy sin'/?= b\ 

RA b 



[18] But sm/3 — 



OR 



1/ a' + b' 

Substituting this value of sin /? into [17] , we get 
[19] 4^y = a'-\-b\ 



2 TO 



[20] and 



ANALYTIC GEOMETRY 



x'y' 



ISIow since x^ and j/ are the coordinates of any point on the 
hyperbola, this must be the equation of the hyperbola. 

by § 39- 

If the asymptotes are the onh^ axes used in any discussion, 
we may drop the accent marks and write the equation 



[21] 



xy- 



Q. E. D. 

275. Corollary. — Since ^ and b maybe any constants whatever 

a'^ b' ^ 

may be any constant. 

4 ■ ^ 

Let this constant be represented by c. 

Then the equation of the hyperbola may be written 

xy = c. 



PROPOSITION XXXII 

276. When the asymptotes are taken as the axes of coordinates 
the equation of the tangent to the hyperbola is 

X y 
in which x' and y' are the coordinates of the point of tangency. 




N ^X 



Fig. 



ASYMPTOTES 2[i 

Let IvT and RS be the asymptotes of the hyperbola. 

Let LT be the Y axis and RS the X axis. 

Let PM be a tangent to the hyperbola at any point P. 

Let x' = OK andy = PK. 
We are to prove that 

4 + 4=2 

X y 
is the equation of the tangent PM. 

Let PN be a secant cutting the hyperbola at the two points 
P and P'. 

Let jc" = OH and y" = P'H. 

Since the secant'PN is a straight line passing through the 
two points P and P', its equation must be of the form 

[i] y__y=-J^|(^'_^). by §58. 

Since the point P is on the hyperbola, its coordinates x' 
and y' must satisfy the equation of the hyperbola. by § 40. 

2 I 7 2 

[2] Hence x'y' = by § 274. 

4 

For the same reason x" andy", the coordinates of the point 
P', must satisfy the equation of the hyperbola. 

b}^ § 274. 



L3J 


Hence 


■^ 4 


[4] 


Hence 


x"y" = xy, 


'.5. 


and 





Substituting this value of y" into [i], we get 
[6] y—y^— _ > {x'—x) 

y' 

[7] and y —y=—^ {x'—x). 
Now let the point P' move along the curve towards P. The 



212 



ANALYTIC GEOMETRY 



secant will continually approach the tangent, and when P' 
reaches P the secant will coincide with the tangent and 

[8] x'^ — x'. 

Substituting this value of x" into [7], we get 



[9] 



y —y 



X 



fio] Hence i =V= — iH 7 

y -^ 



[11] or 



X 

X' ^ y 



Since the x and j/ of this equation are the coordinates of any 
point on the tangent PM, it must be the equation of the tan- 
gent, by § 39. 

Q. K. D, 



PROPOSITION XXXIII 



277. The segment of ayiy tajigent to an hyperbola lying be- 
tween the asymptotes is bisected by the point of tangency. 




Fier. 99 

Let LQ and RS be the asymptotes. 

Let LQ be the Y axis and RS the X axis. 

Let MT be any tangent to the hyperbola at the point P. 

We are to prove that 

PM = PH. 



ASYMPTOTES 213 

Draw PK || YY'. 
Let x' = OK SLudy = PK. 
The equation of MT is 



[I] 




x' ^ y -^• 


by § 276. 


[2] 


Hence 


OH = 2x' — 2OK. 
Since PK II OM, 


by § 45- 


[3] 




OK : OH :: PM : MH. 


by Geom. 23. 


[4] 


And 


OK : 2OK :: PM : MH, 


by [2; . 


[5] 


or 


j_PM 
'~MH' 




[6] 


Hence 


MH = 2PM, 




y. 


and 


PM = PH. 





Q. E. D. 

278. Corollary . — The product of the segments of the asymptotes 
between the ceiiter and any tangent is equal to the sum of the 
squares of the sem,i-axes. 
For 
[i] OM = iy\ by § 276 and § 46. 

[2] and OH = 2X^ . by § 45. 

[3] Hence OM.OH = \x'y\ 

[4] and OM.OH = ^' + ^'. by § 274. 



PROPOSITION XXXIV 

279. The area of the triangle formed by any tangent to an hy- 
perbola, and the segtnents of the asymptotes between this tangent 
and the center is equal to the rectangle of the semi-axes. 

Let LT and RS be the asymptotes. 
Let LT be the Y axis and RS the X axis. 
Let MH be any tangent to the hyperbola. 
Let a = OA and b = OB. 



214 



ANALYTIC GEOMETRY 




Fig. 100 

We are to prove that 

the area of MOH = ab. 

Let (i9=^D0A. 

DA b 



[I 

[2 

[3 
[4: 

[5 
[6 
[7 
[8 

[9 
[10: 



Then 

or 

Hence 
and 

Hence 



tan ce? ^ 



O A 

sin OD b 

a 



a 



cos 00 

a sin CO ■= b cos gl?, 
a^ sin^ Gl> = ^^ cos" ooz=zb^ — b~ sin^ oo. 



by Trig. 3, 
by Trig. 6. 



sin OD = 



b' 



Therefore i — cos 2<i9 1= 
Hence cos 200=: i — 
and cos^ 200 ^^ 

Hence sin" 2Go =z i — 
and sin 200 =z 



a'+b'' 

2b' 



a' + b'' 

b' a' — b' 



by Trig. 5, 



by Trig. 18. 



a'+b'~ a'+b'' 

{a' — b'y 

(a' + b'y 

(a' — b'y 



4a'b' 



{a' + b'y {a' + b'y 
2ab 



a' + U'' 



ASYMPTOTES 215 

[11] But 2c«9=:DOE. by Geom. 15. 

Hence from [10] we get 

[12] sin DOE = ^a I ^g . 

[13] Now the area of MOH — ^OM.OH sin DOE, 

by Trig. 31. 
[14] and OM.OH — a^ ■\- b\ by § 278. 

Hence substituting the right hand members of [12.] and 
[14] into [13], we get 

[15] the area of MOH = i(«'+^"0 ^^ j. =ab. 

Q. K.D. 



CHAPTER XII 



The Parabola 



280. The Parabola. — The parabola is the locus of a point 
moving in a plane in such a way that its distance from a 
fixed point in the plane and its distance from a fixed line in 
the plane are always equal to each other. 




Let F be a fixed point and DG a fixed line, and let P be a 
point moving in the plane DFG in such a way that PF and 
PH shall always be equal to each other. 

Then the line PAM traced out by P is -called a parabola. 

281. The Focus. — The fixed point is called the focus. 

282. The Directrix. — The fixed line is called the di- 
rectrix. 

283. The Focal Radius. — The distance from the focus to 
the moving point is called the focal radius of that point. 

284. The Axis. — The axis of the parabola is a straight line 
drawn through the focus perpendicular to the directrix. 

285. The Vertex. — The point where the parabola cuts the 
axis is called the vertex of the parabola. 



THE PARABOLA 



217 



PROPOSITION I 

286, To draw a parabola which shall have a given point for 
its focus and a given line for its directrix. 

FIRST METHOD ; BY THREAD AND RULER. 




Fig. 102 



lyCt F be the given focus and DR the given directrix. 

Let LMO be a ruler in the form of a right triangle. 

Take an inelastic thread whose length is equal to LO. 
Fasten one end of it at L and the other end at F. Press the 
thread against the ruler at P by the point of a pencil so as 
always to keep it stretched. 

Now move the ruler so that its side MO shall slide along DR. 

The line PAN traced out by the pencil point will be a para- 
bola. 

For at every point P on this line we shall have 

[i] LP + PO = I.P + PF 

[2] hence PF = PO. 

Therefore PAN is a parabola. . by § 280. 



2l8 



ANALYTIC GEOMETRY 

SECOND METHOD ; BY POINTS. 




Let F be the given focus, and DK the given directrix. 
Let XX' be the axis. Take any point A on the axis and 
draw the perpendicular KAK'. 

With AE as a radius and F as a center, draw a circle cut- 
ting KK' in the two points P and P'. 

P and P' will be on the parabola required. 
For draw PG _L DE and join F with P. 
[i] Then PG = EA. by Geom. 17. 

[2] But EA == FP. by construction. 

[3] Hence PG = FP. 

Therefore P is on the parabola. by § 280. 

Similarly it may be shown that P' is also on the parabola. 
By drawing perpendiculars through B, C and H we can de- 
termine other points R and R', S and S', T and T', by the 
same method by which we determined P and P', and can 
show that they are also on the parabola. 

By taking a sufficient number of perpendiculars we can de- 
termine as many points on the parabola and points as near to 
each other as we please. By joining these points we get the 
parabola required. 



THE PARABOLA 



219 



PROPOSITION II 
287. The equation of the parabola is 

j>/' = 2px, 
in which p represents the distance from the focus to the directrix. 




Fig. 104 

Let F be the focus, DE the directrix, and P any point on 
the parabola. 

Let X = OK, y ^ PK, and^ = KF. 

We are to prove that 

y^ = 2px 

is the equation of the parabola. 

2 "2 2 

[i] Tf = fk + pe:. byGeom.26. 

[2] PF = PD = KK.by §280, andGeom. 17. 

[3] Hence PF = EK = /> + FK. 

Substituting this value of PF into [i], we get 



[4] 



(/ + FK)' = FK-f-PK. 



[5] Hence /' + 2/FK + FK = FK + P"K, 

[6] and / + 2/)FK==y. 

[7] But FK = OK — OF = x — OF, 

[8] and OF = OE = i/>. by § 280, 



220 ANALYTIC GEOBIETRY 

[9] Hence FK = ^ — \p. 

Substituting this value of FK into [6] , we get 
[10] p^ -\- 2p{x -- \p) =y\ 

[11] Hence p"^ -\- 2px — p^=y-^ 

[12] or J^' "= 2p-^- 

o. E. D. 

288. Corollary I. — The squares of the ordinates of any two 
points on a parabola are to each other as their abscissas. 

289. Corollary 2. — Ordinates at equal distances from the ver- 
tex are equal to each other. 

290. The Principal Parameter. — 'X\i^ principal parameter 
of a parabola is the double ordinate that passes through the 
focus, the axis of the parabola and the tangent at the vertex 
being the axes of coordinates. 

291. Corollary i. — The pri7icipal parameter of a parabola 
is equal to twice the distance fro7n the focus to the directrix. 

In Fig. 104 let RS be the parameter. 

[i] RF == RM. by § 280. 

[2] RM = EF =p. by Geom. 17. 

[3] Hence RF=:/). 

[4] FS == RF. by § 289. 

[5] Hence FS=/), 

[6] and RS = RF + FS = 2p. 

292. Corollary 2. — The principal parameter of a parabola is a 
third proportional to any abscissa and its corresponding ordinate . 

Since in Fig. 104 the point P is on the parabola, its coordi- 
nates must satisfy the equation of the parabola. by § 40. 

Hence letting the x and y of that equation stand for the 
coordinates of the point P, we get 



]l" 




y^ = 2px. 


by § 287. 


\2\ 


Hence 


X : y :: y : 2p. 


by Geom. 56. 


[3] 


But 


2p = the parameter. 


by § 291. 


[4] 


Hence 


X : y :: y : Parameter. 





Q. K. D. 



THE PARABOLA 221 

PROPOSITION III 

293. The equation of the tangent to the parabola is 

in which x' and y' are the coordinates of the point of tangency 
and p is half the principal parameter . 

For proof compare § 134. 

P 

294. Corollary i. — The fraction ~ is the slope of the tangent. 

295. Corollary 2. — The vertex of a parabola bisects the sub- 
tangent. 



PROPOSITION IV 

296. The equation of the normal to a parabola is 

y , 

.y~y=—— (x'—x), 

in which x' and y' are the coordinates of the point of tangency, 
and p is half the principal parameter . 

For proof compare § 143. 

y 

297 . Corollary i . — The fraction — -=^ is the slope of the normal, 

298. Corollary 2. — The subnormal of a parabola is constant 
and is equal to half the parameter. 

299. A Diameter, — A diameter of a parabola is a straight 
line drawn from any point on it in the positive direction par- 
allel to the axis of the parabola. 



222 ANALYTIC GEOMETRY 

PROPOSITION V 

300. (a) The tangent to a parabola bisects the angle between 
the focal radius of the poijtt of tangency and a diameter produced 
through the point of tangency. 

{b) The normal bisects the angle between the focal radius of 
the point of tangency and a diam^eter drawn from, the point of 
tangency. 




Fig. 105 

Let PT be a tangent to the parabola at P, JH a diameter 
produced through P, and FP the focal radius of the point P. 
We are to prove that 

^fpt = z:hpt. 

Let x' = OK and _y' = PK. 
Let p ^ EF and DR be the directrix. 

FP = PH. by § 280. 

PH = EK. by Geom. 17. 

FP = EK = EO + OK. 

EO = VO—\p. by § 280. 

FP = i/> + ^'. 

FT = FO + OT. 

OT = OK = ^'. by §295. 

Hence by [4] and [7] , [6] becomes 
[8] FT = i/ + ;r'. 



' 1_ 




\2\ 


But 


[3] 


Hence 


[4] 


But 


[5] 


Hence 


\6'_ 




'j\ 


But 



THE PARABOLA 223 

Hence from [5] and [8] , we get 



[9] 




FP 


= FT. 




[10] 


Hence 


^FPT 


= /FTP. 


by Geom. 16. 


["] 


But 


/^FTP : 


= /HPT. 


by Geom. 7. 


[12] 


Hence 


^FPT : 


= /HPT. 





Q. E. D. 

Again let PN be the normal to the parabola at the point P, 
and PJ a diameter drawn from the point of tangency. • 

We are to prove that 

/FPNr=/NPJ. 

[13] /TPL=:/TPN. by §142. 

[14] /HPT=/:FPT. by [12]. 

[15] Hence /HPL=/FPN. 

[16] But /HPL=/NPJ. by Geom. 4. 

[17] Hence / FPN = / NPJ. by Geom. \\ 

Q. E. D. 



PROPOSITION VI 

301. When any diameter and the tangent at its extremity are 
taken as the axes, the. equation of the parabola is 

y- =z 2p X, 

in ivhich p' is the principal parameter divided by the square of 
the sine of the inclination of the tangent. 

Let LN be any diameter and TR the tangent at its ex- 
tremit^^ 

Let LN be taken as the new X axis and TR as the new Y 
axis. 

Let P be any point on the parabola, and draw PM || VY', 
Let X = LM andji/ = PM. 

Let/>'= . /^p.. 
sm LRA 



224 



ANALYTIC GEOMETRY 




Fisr. iq6 



We are to prove that 

y^ = 2p'x 

is the equation of the parabola. 

lyet x' = A J and y ^ I,J, 
y = AK " y'=PK. 

[i] ;»;" = AK= AJ+IvM + MN = ^' + ^+jj/cosPMN. 

by Trig. 2. 

[2] But z^PMN==^I.RA. byGeom. II. 

[3] Hence Ji/' = jr' + .r +ji/cos LRA. 

[4] 
y z=PK=I.J+PN=y + >/sin PMN=y + ysin LRA. 

by Trig, i and Geom. 11. 

Since x" and y" are the coordinates of any point on the 
parabola, when AK and AS are taken as the axes of coordi- 
nates, the equation of the parabola will be 

[5] y = 2px". by § 287. 

Now substituting the values of .r" and j/" found in [3] and 
[4] into [5] , we get 



THE PARABOLA 225 

[6] (y + j/sin LRA)'i=2/'(x'+;tr + j/cos LRA.) 
[7I Hence 
y sin' LRA4-2j'(y sin LRA— /cos LRA)+y'— 2/>;ir'=2;^;»r. 

[8] tan LRA = ^ , by § 294. 

r -. sin lyRA ;^ , /^ • ^ 

[9] or tVt = ^- by Trig. 6. 

'-^-' cosIyRA y ^ £> 

[10] Hence j' sin lyRA — / cos LRA ^ o. 

Since the point L is on the parabola, its coordinates x' and 
y must satisfy the equation of the parabola. by § 40. 

Hence substituting them for the x and y in the equation of 
the parabola, we get 

[11] y^ = 2/^'. 

[12] Hence jk'^ — 2px'=^o. 

Substituting the right hand members of [10] and [12] into 
[7], we get 

[13] y sin^ lyRA = 2px. 

[ 14] Hence y' = -Jf^ . x. 



Now since 



P 



sin' LRA 

,- P 
sin%RA 



[14] becomes 

[15] y=:2/jr. 



Q. E.D. 



302, Corollary i . — Ordinates drawn to the sa?tie point on any 
diameter are equal. 

For since 

[i] y^^2p'x, by §301. 



[2] then y^^-[/2p*x. - 

Now [2] shows that for every positive value of x there are 
two values of jk numerically equal but with opposite signs. 



22Q ANALYTIC GEOMETRY 

303. Corollary 2. — Every diameter bisects a system of chords 
parallel to the tangent at its extremity. 

Compare § 159. 

304. The Parameter of any Diameter. — Th^ parameter of 
any diameter is a third proportional to an}^ abscissa on that 
diameter and its corresponding ordinate. 

305. Corollary i . — The parameter of any diameter is equal to 
the principal parameter divided by the square of the sine of the 
inclination of the tangent at the vertex of the diameter. 

For by [14] of § 301 

[2! Hence x : y :: y : . ., \. ^ . . by Geom. 56. 

2 p 
Therefore . . f^ . = the parameter of the diameter LM. 
sm LRA 

by § 304. 

306. Co7vllary 2. — The parameter of any diameter is four 
times the distance from, the focus to its extremity. 

For in Fie:. 106 



[I 
[2 
[3 
[4 
[5 
[6 

[7 

[8 

[9 
[10: 



FI.= FR. by §300, [9]. 

But FR r^ AF + AR, 

and AR = AJ. by § 295. 

Hence FR = AF + AJ. 

Hence by [i] FL = AF+ AJ = i/ + y. by § 280. 
Again y^=^2px'. by §287. 

Hence x =^ =^. 

2p 

i) 
tanLRA=V. by § 294. 

P 
hence y' — ^^^ \-^p^ =/>cotIvRA, by Trig. 9. 

and y =:;^-cot'I.RA. 



THE PARABOLA 



11'] 



Substituting this value of y'^ into [7], we get 
[11] ;r' = i;^cot'LRA. 

Substituting this value of x^ into [5], we get 
[12] FL^i/' + i/^cot^LRA, 

[13] or Fl. = i/(i + cot^I,RA). 

[14] hence FL := \p cosec^ LRA. 

[15] Hence Fl. — ^ 



2 sin' IvRA 
[16] Therefore Parameter 



2p 



sin' IvRA 



by Trig. 10. 
by Trig. 7. 

4FI,. by §305. 
Q. E.D. 



PROPOSITION VII 



307, The squares of ordinates to any diameter of a parabola 
are to each other as the cor^'esponding abscissas. 




Fig. 107 

Let IvN. be any diameter and TR a tangent at its extremity. 
Let LN be taken as the X axis and TR as the Y axis. 
Let PM and P'N be any two ordinates drawn to the diame- 
ter LN. 

Let X = LM and y = PM. 
^'=LN '' y=P'N. 



228 ANALYTIC GEOMETRY 

We are to prove that 



2 /2 

y ■ y 



X : X 



Since the point P is on the parabola, its coordinates must 
satisfy the equation of the parabola. 

by §301. 

by § 301. 



[I] 


Hence 




J/^ = 2p^X. 


~_2 


Similarly 




y = 2fx'. 


'.'h. 


Hence 




y X 
y'~~ x" 


A 


or 


y 


: y^ \: X \ x' 



Q. K. D. 



PROPOSITION VIII 

308. When any diameter and the tangent at its extremity are 
taken as the axes of coordinates, the equation of any tangent to 
the parabola is 

_ P 



y —y 



T (^'--^), 



in which x' and y are the coordinates of the point of tang en cy of 
the latter tangent, and p' is equal to half the principal parameter 
divided by the square of the sine of the ijiclination of the tangent 
taken as the Y axis. 




Fig. 108 



THE PARABOLA 229 

Let LM be any diameter and RS the tangent at its ex- 
tremity. 

Let LM be the X axis and RS the Y axis. 

Let PT be a tangent to the parabola at any point P. 

Let ^' = LK and y = PK, 
andlet/ = ^-^|^^. 

We are to prove that 

y— jj/=y (x'—x) 

is the equation of the tangent to the parabola. 

Let PP' be a secant cutting the parabola at the two points 
P and P'. 

Let x" = LH and /' = P'H. 

For the method of demonstration see § 221. 



230 



ANALYTIC GEOMETRY 
PROPOSITION IX 



309. The equation of the chord which joins the points of tan- 
gency of two tangents drawn to a parabola from any point 
without it is 

yy' — p{x-\-x'), 

in which x' and y' are the coordinates of the point from which the 
tangents are drawn, and p is half the principal parameter. 




Fig. 109 

Let PT and P'T be any two tangents drawn to the parabola 
from the same point T. 

Let PP' be the chord joining their points of tangency. 

Let x' ^ AR and j/' = TR. 

We are to prove that 

yy' — p{x -\- x') 
is the equation of PP'. 

Let x" = AK and y = PK, 
;t:'" = AH " y" = P'H. 

The equation of PT is 



[I] 

[2] Hence 
[3] and 



y 



y 



■-r = f (-' 



X) 



by § 293. 



— yy" = px" — pX, 
yy" — px +y" —px". 



THE PARABOLA 231 

Since the point P is on the parabola, its coordinates ;f" and 
y must satisfy the equation of the parabola. 

Hence substituting x" and y' for the x andjK of the equa- 
tion of the parabola, we get 

[4] y"^—2px". by §287. 

Substituting this value ofy* into [3], we get 

[5] yy"^P^x^x") 

for the equation of PT. 

Similarly we may show that the equation of P'T is 

[6] yy''^p{x^x'"). 

Now since the point T is on the tangent PT, its coordinates 
x' andy must satisfy the equation of PT. 

Substituting x' and y' for the x and y of that equation, 
we get 

[7] yy'=/(^'+y'). 

Similarly, since the point T is on the tangent P'T, we get 

[8] ^y"=/>(y+y"). 

Now equation 

[9] yy' —p{x-\-x') 

is the equation of a straight line. by § 67. 

But the coordinates ^" andy" of the point P will satisfy this 
equation, for if they are substituted for the x and y in it, we 

get [7]. 

Hence the straight line represented by [9] must pass 
through P. by § 41. 

The coordinates x'" and y'" of the point P' will also satisfy 
[9] , for if they are substituted for the x and y in it, we get [8] . 

Hence the straight line represented by [9] must also pass 
through P'. by § 41. 

Therefore since the line represented by [9] passes through 
both the point P and the point P', that equation must be the 
equation of PP'. q. e. d. 



232 



ANALYTIC GEOMETRY 



PROPOSITION X 

310. If any chord of a parabola pass through a fixed point and 
tangeiits be drawn at its extremities, and if the chord be made to 
revolve about the fixed point, then the locus of the intersection of 
the two ta7ige7its will be a straight line whose equation is 

yy' — p{x -\- x') , 

in which x' and y' are the coordinates of the fixed point, and p 
is half the principal para meter . 




Fi.sr. no 



Let PP' be a chord passing through the fixed point R, and 
let PT and P'T be tangents drawn at its extremities and inter- 
secting at the point T. 

LetJc' = CH and y = RH. 

Let PP' revolve about R as a pivot. 

We are to prove that the locus traced out by T is a straight 
line, and that its equation is 

Let y = CK and j/" ^ TK. 

The equation of the chord PP' is 

[i] jj/jf"=/(^ + ^"). by § 309. 

Since the point R is on this chord, its coordinates x^ and y' 
must satisfy the equation of that chord. by § 40. 



THE PARABOLA 233 

Substituting x^ and j/' for the x and _y of [i], we get 
[2] yy=p{x' + x"). 

Now let PP' revolve about R as a pivot. Then T will trace 
out a locus. Moreover [2] will be satisfied by the coordi- 
nates of the point T wherever it may be, as it traces out this 
locus. 

Hence x^^ andj/" of [2] stand for the coordinates of every 
point on the locus traced out by T. 

Therefore [2] must be the equation of that locus, by § 39. 

This locus is a straight line. by § 67. 

Since T stands for any point on this straight line traced out 
by the intersection of the tangents, we may drop the accent 
marks from its coordinates and write them x and j/. Hence [2] 
may be written 

jy' — p{x-{- x'). 

Therefore the locus is a straight line and its equation is 

j/y' =ip{x-\- x') . 

Q.E.D. 



PROPOSITION XI 

311. If two tangents he drawn at the extremities of any focal 
chord of a parabola 

(i) the tangents will meet on the directrix ; 

{2) the li7ie joining the hitersection of the two tajigents to the 

focus will be perpendicular to the focal cho7^d. 

L<et PT and P'T' (I^ig- m) be the two tangents drawn to 
the parabola at the extremities of the focal chord PP', and let 
R be the intersection of the two tangents. 

Let DD' be the directrix. 

We are to prove 

(i), that R will be on the directrix DD', and 
(2), that RF will be perpendicular to PP'. 



234 



ANAL YTIC GEO ME TR Y 




Fig-. Ill 



Let x' = AK, y = RK, and/= FK. 
The equation of PP' is 
[i] yy' ~ p{x' -^ x) , by § 309. 

in which x' and y' are the coordinates of the point R and x 
and y are the coordinates of any point on the chord PP'. 

Since the point F is on the chord PP', its coordinates 
x:=^\p and j/= o must satisfy the equation of the chord. 

by § 40. 
Substituting these values for the x andjF of [i], we get 
[2] o^p{x'^\p). 

[3] Hence ^' = AK = — \p. 

But — \p is the distance from the vertex of the parabola to 

the directrix. . by § 280. 

Therefore the point R is on the directrix, Q- E. d. 

Since RF is a straight line passing through the two fixed 
points R and F, its equation must be of the form 



[4] 



y'-y~y 



y 



-,^x'~x). 



by § 58. 



X X' 

Now in [4] let x" and j/" stand for the coordinates of the 
point F, and x' andy the coordinates of the point R. 

Then x" = ip and y = o. 

Substituting these values into [4] , we get 



THE PARABOLA 235 

[5] y-y = .^^^Ax-~x). 

[6] But x" — —\p. by [3]. 

y' 

[7] Hence y — ^= — —-( — \p — x). 
[8] -y = -^{-hp-x)-y\ 

[9] ■ y^^k-\p-x)^y'. 

[10] ^=_z,+z. 

which is the equation of the line RF. by § 39. 

From [3] we get 
[11] :r' = AK = — i/. 

Substituting this value of x' into [i], ^ve get 

[12] yy '=^p{—\p-\- x). 

[13] j^ = -ya/-^), 

[14] y ~ ^^ — —,^ 

L -rj y ^y 

which is the equation of the chord PP'. by § 39. 

Now let s = the slope of RF, 
and / =. the slope of PP'. 
Then from [10] we get 

and from [14] we get 

[16] ^' = 4- 

[17] Hence ss' := — "— ^ = — i. 

P y 

[18] Hence i + ^5'=o. 

Therefore RF is perpendicular to PP'. by § 62. 

Q. E. D. 



236 



ANALYTIC GEOMETRY 



PROPOSITION XII 



312. The two tangejits drawn at the extremities of any f oca t 
chord a7'e perpendicular to each other. 




Fio'. 112 



Let PT and P'T' be the two tangents drawn at the ex- 
tremities of the focal chord PP'. 

Let R be the point on the directrix at which the two tan- 
gents meet. 

We are to prove that PT is perpendicular to P'T'. 

Let x' = AK and / = RK. 

" y' = PM, 



x"^^AM 

x'" = AN 



y 



P'N. 



Since PT is a straight line passing through the two given 
points P and R, its equation must be of the form 



in vc^hich 
[2] 



y—y='^, — ^j(y—^) 



X 



■X 



y -y 



X' 



■X' 



the slope of PT. 



by § 58. 



by § 59. 



THE PARABOLA 237 

[3] But A= the slope of PT. by i .94. 

[4] Hence ''^•^ ^- 

Since the point P is on the parabola, 
[5] y'- — 2px". by § 287 and §40. 

[6] Hence x"—^^, 

2p 

[7] Also x' ——\p. by §311, [3]. 

Substituting the right hand members of [6] and [7] into 
[4] we get 



[8] 



2 p 2 



[9] Hence j/"=— yy z=-^ + ^, 
[10] y^' — 2y'y"—p\ 



[11] and' ■ y' = VP"-\-y'" + y'' 

p p 

[12] Hence ^ = ; = the slope of PT. by [3]. 

y y p +y -\-y 

Since P'T' is also a straight line passing through the two 
given points P' and R, its equation must be of the form 

[13] y-j' = ~;^"iZ;^,' (^'-^), by §58. 

in which 

[14] ~^n ,Z;^^ = ti^^ s^^p^ ^^ p"!^'- ^y § 59. 

[15] But —^n— the slope of P'T'. by § 294. 

y 

r -1 —y"'—y p 

[16] Hence — yT,"^:^, = — yr,- 

Solving this equation as we did [4], we get 

[17] y"' = Vf + y"--y'- 



238 



ANALYTIC GEOMETRY 



[i8] 


Hence 


P'T'. 






Now 




and 



p_ _ p__ 



- = the slope of 



by § 15. 



[19] Then ss' = — 



let ^ = the slope of PT, 
s' = the slope of P'T' 

P 



X 



/ 



^ p' +y' — y v^^' +y' +y 






P'+y'-y 

[20] Hence i -|- /^ r= o. 

Therefore PT and P'T' are perpendicular to each other. 



by § 62. 
Q. E. D, 



PROPOSITION XIII 



313. When the focus is take?i as the pole, the polar equation 
of the paiabola is 

P 



I — cos 



a ' 



in which p is half the principal parameter, r is the radius vector 
of any point 071 the parabola, and is its vectorial angle. 



t 







1 


}/ 


u 


1 


Y 


1 








/ 






E 


i 








\ 


F 


K 




H 


k 


\ 



Fig. 113 



lyCt the focus F be the pole. 



THE PARABOLA 239 

Let P be any point on the parabola, r its radius vector, 
and its vectorial angle* 

Let/ = one half the principal parameter HH'. 
We are to prove that 

P 
^^ ^ ' 

I cos C7 

is the equation of the parabola. 

Let DD' be the directrix. 

DrawPDJ_DD' and PK_LXX'. 

[i] FP = PD. by §280. 

[2] But PD = EK. by Geom. 17. 

[3] Hence FP = EK = EF+ FK, 

[4] also EF =/>, by § 291. 

[5] and FK = r cos Q. by Trig. 2. 

[6] Hence by [3] FP^/+rcosO, 

[7] or r =^ p -\- r Qos d . 

[8] Hence r(i — cosO) =z p. 



[6] Therefore 



— cos 



Q. K. D. 



EXA.MPLES 

1. What is the value of r when = o? Ans. r = oo. 

2. What is the value of r when =^ 90° i* 

Ans. r = / = FH. 

3. What is the value of r when d = 180° ? 

Ans. r—^ — AF. 
2 

4. What is the value of r when d = 60° ? 

5. What is the value of r when Q = 270"^? 



CHAPTER XIII 

The Conic 

314. The Conic. — A conic is the locus of a point moving in 
a plane in such a way that its distance from a fixed point in 
the plane, and its distance from a fixed line in the plane 
always have the same ratio to each other. 

315. The Focus. — The fixed point is called W\^ focus oiWv^ 
conic. 

316. The Directrix. — The fixed line is called the directrix 
of the conic. 



PROPOSITION I 

317. The ellipse is a conic. 

For by § 176 the ellipse is only a particular case of the locus 
defined in § 314. 

318. Corollary. — The ellipse is a conic in which the distance of 
any poifit on the conic from the focus is less than its distance from 
the directrix. 

Compare §'176 and § 129. 



PROPOSITION II 

319. The hyperbola is a conic. 

For by § 251 the hyperbola is only a particular case of the 
locus defined in § 314. 

320. Corollary. — The hyperbola is a conic in which the dis- 
tance of any point on the conic from the focus is greater tha7i its 
distance from, the directrix . by § 252. 



THE CONIC 241 

PROPOSITION III 

321. The parabola is a conic. 

For by § 280 the parabola is onl}^ a particular case of the 
locus defined in § 314. 

322. Corollary. — The parabola is a conic in which the distance 
of any point on the conic from the focus is equal to its distajice 
fro7n the directrix. h\ § 280. 

Poles and Polars 

323. Equations [5], §255^, [9], §256, and [4], §261, and 
the similar equations of the other conies show that the coor- 
dinates of every point on the tangent and the coordinates of 
the point of tangencA^ ; the coordinates of every point on 
the chord of contact and the coordinates of the intersection of 
the corresponding tangents ; and the coordinates of every 
point on the locus of the intersection of two tangents, and 
the coordinates of the fixed point about which the corres- 
ponding chord of contact revolves, are connected in the same 
wa}^ 

That is, each of these three lines is connected with a fixed 
point in the same wa}' in which each of the other lines is con- 
nected with a fixed point. 

324. The Polar. — If the coordinates of every point on any 
line have the same relation to the coordinates of a certain 
fixed point that the coordinates of every point on a tangent 
have to the coordinates of the point of tangency, that line is 
called the polar of the fixed point. 

325. A Pole. — The fixed point is called X\\^ poleoi the line. 
When the pole is without the conic the polar is the chord of 

contact of the two tangents drawn from the pole. 

When the pole is on the conic the polar is a ta7igent to 
the conic. 

When the pole is within the conic the polar '\'$, the locus of 
the i7itersection of the two tangents drawn at the extremities 
of the chord of contact passing through the pole. 



242 



ANALYTIC GEOMETRY 



PROPOSITION IV 

326. In an hyperbola or an ellipse the polar of a point on any 
diameter or on any diameter produced is parallel to the conjugate 
diameter. 

I. FOR THE HYPERBOI^A. 




Fig. 114 

Let P be any fixed point on the diameter lyM produced, and 
let ST be the polar of that point. 

Let NO be conjugate to LM. 

We are to prove that ST is parallel to NO. 

Let x' = CK and j/' = PK, 

Q = ZlMCK " 0' = ^NCK. 

Since LM is a straight line passing through the two points 
C and P, its equation must be of the form 



[i] 



7 — y 



by § 58. 



X X 

in which x" andy" are the coordinates of the point C, and x' 
SLudy are the coordinates of the point P. 

Since C is the origin, 

x'' = o and jk" = o. 
Substituting these values into [i], we get 



[2] 



y—y—~{x'—x) 



X 



[3] Hence y'x' — yx'=^y'x' — jj/'jr, 



THE CONIC 



243 



[4] and 



y 



_y 



X' 



X, 



which is the equation of LM, 
[5] Hence 
Since IvM and ^O are conjugate to each other, 



^ = tan 0. 

X 



[6] 



tan 6 tan 6' = 



a' 



by § 55 



by § 240. 



y 

fyl Hence — ;^tan0'=^ 

^'■^ X a' 



[8] Hence 



tan S' = 



'?'x^ 

?y 



The equation of the polar ST is 
[9] a^yy — U'-xx'^^ — a'b^. by § 324 and § 221 

[10] Hence 
[11] Hence 



y 



b'x' 

-2 ./ 



X — 



a y 
tan SRK = 



y 

b' X^ 
2 V 

ay 

By comparing [8] and [11], we get 
[12] tan Q' = tan SRK. 

Therefore ST is parallel to NO. 

2. FOR THK ELIvIPSE). 

Y vS 



by § 53- 



by Geom. 9, 
Q. E. D. 




Fig. 115 



For demonstration compare that for the hyperbola. 



244 



ANALYTIC GEOMETRY 
PROPOSITION V 



327. For a pai^abola the polar of a point on any diameter is 
parallel to the ordinates of that diameter. 




Fig. 116 

Let P be aii}^ fixed point on the diameter LM, and let ST 
be the polar of that point. 

Let NK be an ordinate to the diameter LM. 

We are to prove ST |i NK. 

Let LO be a tangent to the parabola at L- 

Let LM be taken as the X axis and LO as the Y axis. 

Then the equation of ST will be 



[i] 



,._,= .,._.,, .,.3.....,3. 



[2] Hence yy :^ pi^x^ -\- x) . 

Now since the point P is on the X axis, jj/' = o and [2] 
becomes 

[3] x = —x\ 

But this is the equation of a line parallel to the Y axis. 

by § 'j^a. 
Hence ST || LO. 

But NK II LO. by § 14. 

Therefore the polar ST [|, the ordinate NK. 



THE CONIC 



245 



328, Corollary. — The polar 0/ any poi?it on an axis of a conic 
is perpe7idicular to that axis. 



PROPOSITION VI 



329. In an hyperbola and in an ellipse the distance from the cen- 
ter measiired along ajiy diameter to the pvlar of any point on 
that diameter or diameter produced , is a third proportio7ial to the 
scTni-diameter and the distance of the point fvm the center. 



I. FOR THE HYPERBOLA. 




Fig. 117 

Let ST be the polar of the point P on the diameter lyM 
produced, and let R be the point where this polar cuts LM. 

Let a' = CM and ^ = CP. 

We are to prove that 

CR : a' :: a' : CP. 

Let NO be conjugate to LM. 

Let LM be taken as the X axis and NO as the Y axis. 

Let ^' = CN. 

Then the equation of the polar ST will be 

[i] a'^yy — b''xx^ ^= — a'^V^ . by § 324 and 255(2. 

Now the polar ST cuts the X axis LM where y ^o and 
-r=CR. by §45. 



246 



ANALYTIC GEOMETRY 



Substituting these values into [i], we get 
[2] — /^'VCR = — ^'-^^'^ 



[3] and 
[4] Hence 
[5] or 



CR 



a 



X' 



CR : a' :: a' : x\ 
CR : a' :: a' : CP. 

2. FOR THE ELIvIPSE. 



Q. E. D. 




Fig. 118 
t 

lyet SR be the polar of the point P on the diameter LM, 
and let R be the point where this polar cuts LM produced. 

Let a' = CM and x' ^ CP. 

We are to prove 

CR : a' :: a' : CP. 

For demonstration compare that of the hyperbola. 



PROPOSITION VII 

330. T/ie distance from the extremity of any diafneter of a 
parabola measured along that diameter to the polar of any point 
on it is equal to the distance from, the vertex to the pointy but is 
measured in the opposite direction. 



THE CONIC 247 

For in § 327 we have proved that 

in which x is the abscissa of any point on the polar, and x^ is 

the abscissa of its pole. 

o. E. D. 



PROPOSITION VIII 
331. In any conic the directrix is the polar of the adjacent focus. 

I. FOR THE HYPERBOLA 





Fig. 119 

Let DD' be the directrix and F the adjacent focus. 
We are to prove that DD' is the polar of the focus F. 

Let « = CA and p = RF. 

[i] Then CR = CF — RF. 

[2] But - CF = ^=^^, by §218, [2] 

a{e'^ — i) 



[3] and 



RF =/> 



by § 251, [5] 



[4] Hence by [i] CR= ae — 



a(^e^ — i) 



The directrix _L the X axis. 
The equation of the polar of the focus is 



by § 248. 



248 ANALYTIC GEOMETRY 

[5] / — r = -^/(-r' — x) , by § 324 and §221. 

a'y 

[6] Or a'yy' — U'xx' = —a'b\ 

in which x^ = ae and_>'' = o. 

Substituting these values into [6], we get 

[7] — y^xae = — a^d"^, 

a 



[8] and x 



e 



which is the equation of a line parallel to the Y axis, and at 
a distance from the origin equal to — . by § 56. 

The polar of the focus is therefore perpendicular to the X 
axis. by Geoni. 6. 

Since the origin is at the center, the distance of the polar 

of the focus from the center is — . 

e 

Hence from [4.] and [8] it follows that the directrix and 
the polar of the focus are at the same distance from the cen- 
ter and both perpendicular to the X axis. 

Therefore the directrix coincides with the polar of the focus. 

by Geom. 2. 
Q. E.D. 

2. FOR THE ELTvIPSK 

For the demonstration of the ellipse compare that of the 
hyperbola. 

3. FOR THE PARABOIvA 

lyCt DD' (Fig. 120) be the directrix and F the focus. 

We are to prove that DD' is the polar of the focus F. 

[i] AR==AF. by §280. 

The distance of the polar of the focus from A is equal 
to AF but measured in the opposite direction from A. 

by § 330. 




249 



Fig. 120 

Therefore the directrix and the polar of the focus pass 
through the same point R. 

The directrix is perpendicular to the X axis. by § 284. 

The polar of the focus is perpendicular to the X axis. 

by § 327 and § [4. 

Therefore the directrix coincides with the polar of the 
focus. by Geom. 2. 

Q. E.D. 
EXAMPLES; 

1. The slope of a tangent to the parabola j/^ = 6x is 3. 
What is the equation of the tangent ? Ans. J^ = 3-^ + y. 

2. Given the parabola jj/^ = 8 jr. What is the parameter of 
the diameter jF — 16=0? Ans. 136. 

3. Required the chord of contact of tangents drawn from 
( — 2, 5) tojj/^ = 8;t:. Ans. 5j/ — ^x-\-8 = o. 

4. A tangent to j/^ = 4.x makes an angle of 45° with the X 
axis. What is the point of tangency ? 

5. Given the parabola jj/^ = 4-^ ; required the equation of 
the chord which is bisected by the point (2, i). 

6. Required the equation of the right line passing through 
the vertex of any parabola and the extremity of the focal 
ordinate. 



250 ANALYTIC GEOMETRY 

7. The headlights on locomotives contain parabolic reflec- 
tors. Why ? 

8. The equation of a parabola is y^ = lox. Through the 
point (7, ord. -|-) we draw a tangent and a normal. Required 
the lengths of the tangent, normal, subtangent and sub- 
normal. 

9. Thepointsof contact of two tangents are (-^i,Ji), (-^...JKJ. 
Find their intersection. 

10. Find the equation of a straight line touching the parab- 
olajK^= i6jr and passing through ( — 4, 8). 



CHAPTER XIV 

The General Equation of the Second Degfree 



PROPOSITION I 



331 . To determine the forms of all the loci which are repre- 
sented by equations of the second degree co?itaining two variables 
only. 




Fig:. 121 

The general equation of the second degree may be written 

[i] Ax" -\- 2Bxy -{- Cy^ -\- 2Dx -\- 2Ey -{- F=o. 

If the X and y in this equation be variables and represent 
the coordinates of a point moving in a plane, then whatever 
real values its coefficients may have, the equation will repre- 
sent some locus, real or imaginary. 

Let LM of Fig. 121 represent that locus. We are to find 
the forms of LM for all real values oi A, B, C, D, E, F. 

It will be convenient to divide the investigation into tw^o 
parts. 

I St. To find the forms of LM when C f=^* o. 

2nd. To find the forms of LM when C = o. 

* Read ^f; "is not equal to." 



252 ANAL YTIC GEO ME TR Y 

FIRST PART 
C^ O. 

Equation [i] may then be written 

[2] Cf + 2{Bx-^E)y + Ax' + 2Dx + 7^= o. 
Finding the values of jk from this, we get 



[3] 7 



Bx A- E I 

^ ± ~^^/\Bx^Ey—C{Ax'^2Dx^F) 



C 



[4] Let R^{Bx + EY — C{Ax' + 2Z?jr + F) . 

[5] Then 

^ = BV + 2i?i5^;r + E' —ACx' — iCDx— CF, 
[6] or ieEE(i5'^— ^0-r''+ 2(^9^5-— (ri>)jr+^'^—^Ci^. 

^ ^ i7 



[7] I^et 



^ = + 



B C E 



D E F 

Wentworth's College Alg., § 399. 

The coefficients of [6] may easily be remembered by notic- 
ing that they are the minors corresponding to F, Z^and A in /I. 





A 


B 


B' AC = 








B 


C 




B 


C 


BE~CD = 








D 


E 




C 


E 


E' CF = 








E 


F 



and 



Multiplying both sides of [7] b}^ — C, we get 

[8] —CA — — A CF^ A CE' + CW + CFB' — 2 CBDE. 
Adding B'E' — B'E' io the right hand member of [8], we get 
[9] -CA = 
— ACF+ ACE' — B'E' + CFB' + B'E' — 2CBDE+ CW 
^ ACiE' — CF) — B\FJ — CF)-^ {BE— CDY 
— {AC--B'){E'—CF) + {BE — CDy. 



GENERAL EQUATION OF THE SECOND DEGREE 253 
[10] or — Cz/=r ^BE—CDY— {B' — AC){E' — CF) . 

Ivet a^B- —AC, 

b^BE— CD. 
c = E' — CF. 

Then equation [6] becomes 

[11] R^=^ ax^ -\- 2bx -\- c, 

and [10] becomes 

[12] b' — ac=: — CA. 

Now it will be found that the form of I^M depends largely 
upon the coefficient of x^ in [6]. 

This coefficient may take three different values. 

ist. B' — AC<o. 

2nd. B' — AC>o. 

3rd. B' — AC:^o. 

First Case 

B' — AC < o. 

In this case we may have 

ist. C^ < o, 

2nd. Cz/ > o. 

3rd. €/} = o. 

First when C^ < o. 

[13] Put ax~ + 2bx -\- c ^ o. 

Now since C^ < o, 

—C^ > o ; 
hence by [12] 

b^ — ac ^o. 

Hence the roots of [13] are real and unequal. 

Wentworth's College Algebra, § 141. 

.Let these roots be x' and x" and let x^ < x" , 



254 ANAL YTIC GEO ME TR Y 

[14] then R=^ ax^ -\- 2bx -\- c^=i a{x — x^){x — x"). 

Wentworth's College Algebra, § 153. 

But since in this first case B- — AC <^o, a is negative, and 
[14] may be written 

[15] R — — a{x — x'){x—x'''). 

Substituting R for the quantity, under the -j/ in [3] , we 
o'et 

r . T Bx ■\- E ^ I -7— 
[16] J/ = ^ =±= -^ 1/7?. 

[17] Ivet y = -^T/^ 

[18] then y^—?^^^^Y^ 

which is the equation of LM. by § 39. 

If in Fig. 121 P be any point on LM, then in [18] 
[19] 7=PK and ^ = OK. 

Let AB be the line whose equation is 

r , Bx-\-E 
[20] y— -^ — . 

Let the coordinates of the point Q on this line be 
;i: = OK and /" = QK. 

Rx A- F 
[21] Then /" = c ' by §40. 

Now if we subtract [21 from [18], member for member, 
we get 

[22] y-y">^^Y^ 

[23] or PK — QK = dry, 

[24] or PQ = ± y- 

Take HI = IJ, 

and draw lY' || OY. 

If now we take ooY' and cioX' as our axes of coordinates, 
then Gi?Q and PQ will be the coordinates of the point P. 



GENERAL EQUATION OF THE SECOND DEGREE 255 
\_2\a\ Let 0!9Q = X. 

[24^] PQ= Y. by [24]. 

In Fig. 121 let O be so taken that 

x' = OH and jt" = OJ. 

Then x" — OH + 2HI, 

hence x' ^ x" = 2 (OH + HI), 



[25] and ^^tj^ = 01. 

2 

Hence by [19] and [25], we get 

[26] OK = ;t:= ^ +G^S. 

2 

Now by Trig. 14 and Geom. 11 w'e get 

ce9S _ sin ce^Q S _ sin CAP 
69Q "" sin Ce^SQ ~ sin COA" 

-^ sin CAO . ^^ • ^ • ,7 

But —. — ^^ ■ is constant. Representmo: this constant bv k, 
sm COA 

we get 

[27] — FV = ^^ 

[28] G9S=/^&?Q=^X by [24^]. 

Substituting into [26], w^e get 

x' -4- jf" 
[29] x^ ^ -\-kX. 

2 

Substituting this value of x into [15], we get 

[30] R= — a{ ^^'^ ^" +kX—x')(^ ''''^^ """ + kX—x"y 

[31] whence R = — a (^kX+ '^'^^) (^kX—"^—) . 

x" — x' 

Now let /i = , 

2 

[32] then ^ = — a(^-X' — /z'). 

Substituting this value of R into [17], we get 



256 




ANALYTIC GEOMETRY 

T 


[33; 


Y— V a{k'X' If). 


[34; 




CY'= ak'X' + ah\ 


L35. 




Y' — X'-\- I 
ah' h ^ ' 


[36] 


Let 


a I 

ah' b"' ' 


[37] 


and 


k' I 
h' a"' 


Substituting 


into [35] , we get 


[38] 







[39] Whence a" Y' + b"X' = a"b'\ 

Now since b}^ [24a] and [24*5] A"=: ooQ and Ki= PQ, [39] 
is the equation of the ellipse. 

Second when C/i > o. 
In this case, by [12], b' — ac <Ci o. Hence the roots of 

ax' -\~ 2bx -\-c=^ o 
are imaginary. Wentworth's College Algebra, § 141. 

Hence ax^ -\- 2bx-\- c is negative when a is negative. 

Wentworth's College Algebra, § 180. 
But in this first case 

B''—AC<Co, 
[40] or a < o, 

that is, a is negative. Hence ax^ -\- 2bx -\- c is negative. 

Hence by [11] ^ is negative, and therefore |/ 7? is imagi- 
nary. 

Hence by [16] jj^ is imaginary. 

Therefore when C^ > o, [i] cannot represent any real 
locus. 

Third when CA = o. 

Since in this first part of the investigation C^ o, then z/ = o. 



GENERAL EQUATION OF THE SECOND DEGREE 257 

Hence by [12] 

[41] /^" — ac = o. 

Therefore the roots x' and .r" of [13J are real and equal. 

Wentworth's College Algebra, § 141. 

[42] Hence R=:- a{x — x'){x — x')^^a{x — x'Y, 

and 'Y^V ^= ~7^ (-^ — ^') V ^' 

Hence [16] becomes 

r- -, • Bx -\- E X — x' . — 

[43] y= -^ — ± ^~^^' 

Now since by [40] a is negative, the last term of this equa- 
tion is imaginary, and the equation itself represents two im- 
aginary straight lines. 

Therefore when B'^ — AC <i o, and 

ist. C^ < o, [i] represents an ellipse ; 

2nd. CA > o, [i] represents an imaginary ellipse ; 

3rd. CA = o, [i] represents two imaginary straight lines. 

Second Case 

B'—AC>o. 
In this case again we may have 
ist CA <^o. 

2nd. CA ^ o. 

3rd. CJ r^ o. 

First let CJ < o. 

By [12] we see that when CJ < o, b^ — ac^o. Hence 
again x' and .r" are both real and unequal. Also since 
B' — ^C> o, ^ is positive. Therefore [33] will become 

[44] Y^^y'^k'X'—h'). 

[45] C- r- = ak'X' — ah'. 

[46] _^=^X'-:. 



258 ANAL YTIC GEO ME TR Y 

C I 






lyet 

and 

then [46] will become 

[47] ^ = ^-1. 

[48] and a'^y — V'X' — — a'''b'\ 

which is the equation of the hyperbola. 

Second let C^ >o. Then by [12] b'^ — ac<^o, 
and the roots of [13] are imaginary. 

Now since in this second case dispositive, as in [31], we get 

^" — x\ ( . ,. jr" — x^ 



R^a[ kX+ kX 



2 / \ 2 



but since x' and ^" are imaginary, we ma}^ write this 
R = a{kX-\~ h^'^^i){kX— h y~^^i ) 
— a{k''X^^Jr^. 

Therefore by [17] Y —~y'a{k''X' + h'). 

C'Y'z^ak'X' + ah\ 
C'Y'—ak'X'=^ah\ 



C^ I Jz' I 

Y- X' 

then -jjT, 72 = I ' 

b a 

or a'^Y' — b^'X-' — a'^b'^ 

which is the equation of the hyperbola conjugate to [48]. 

Third let 6^ = o. 

Again b\^ [12], when CA = 0, b^ — ac^=^ o, and the roots of 
[13] will be real and equal; and again as in [42] 



GENERAL EQUATION OF THE SECOND DEGREE 259 

[49] R^^a{x — x'Y, 

and as in [43] 

r- -, Bx -\- E X — x' . — 

[50] r = -^ ± — -^—V a. 

Now since a is positive, this equation represents two real 
straight lines. 

Therefore w^hen R- — AC > o, and 

ist. CA <^ o, [1] represents an hyperbola; 

2nd. CA > o, [i] represents the conjugate hyperbola ; 

3rd. 6^ = 0, [i] represents two straight lines. 

Third Case 

B'—AC=o. 
Since a = B' — AC, 

[51] then « = o, 

and by [6] w^e get 

R— 2{BE— CD)x + E' — CF, 
[52] or by [11] R^^2bx-\-c. 
Substituting this value of R into [3], we get 
Bx-^E , I 



[53] y = ^ ±-^^2bx-^c 



r -, Bx+ E ^ ^ J , ^ \ 



Now the values of y given by this equation will be real so 
Ions: as 



J3 

= c 

X 



> 2b' 
Since by [51] 

<2 = o, 

[55] then {^BE— CDf^-CA. by [10] . 

[56] Hence b''=i — CA, 

so that when ^ — to o, CA is negative, 



26o ANALYTIC GEOMETRY 

[57] and when ^ = o, CA is o. 

Hence in this case we can only have 

ist. CA < o. 

2nd. CA = o. 

First let CA <Co. 

Then by [17] and [52] we get 

[58] r= -^ v^bx^c = ^V ^K ^+ ^) 

In Fig. 121 let 

I 



thenby[58] Y=^ -^ y" 2b{x — x;) 

Now as in [29] we may show that 

;i: = jTj -(- kX. 
I 



[59] Hence K=— ^ ]/ 2<5(;t:^ + /^X — x^), 



[60] and r= — |/ 2^/^X, 

[61] and Y'^2~X. 

Let -^ = ^ ' 

[62] then F^ = 2/X, 

which is the equation of the parabola. 

Second let CA := o, 

Then by [57] 

<^ = o, 
and [53] becomes 

[63] J = 7^ ± 



which is the equation of two parallel straight lines. 



GENERAL EQUATION OF THE SECOND DEGREE 261 

Therefore when B' — AC^=-o, and 

ist. CA << o, [i] represents a parabola. 

2nd. C^ = 0, [i] represents two parallel straight lines. 



SECOND PART 

C=o. 
Equation [i] may then be written 

[64] 

[65] y 



[2{Bx + E)']y+ Ax' + 2Dx -^ F=o. 
Ax'^2Dx-^ F 



2 {Bx + E) 



[66] y 



A __D^ AE 
2B "^ B ^ 2B' + 



AE' , 2DE . ^ 
^i ^^ E 



B'- 



B 



[67] y 



A _ AE 



D 



+ 



2{Bx^E) 
— AE' + 2BDE—EB' 



B 
But when C= o, 

A — —AE'-^2BDE — EB' 
Hence [67] becomes 



2B\Bx^ E) 



[68] y 



A AE 



~B 



+ 



A 



r . A , AE E , 

[69] -^ = -^^+5^-^ + 



2B\Bx+E)' 

A 



2BHx + 



E^ 
~B 




Fisf. 122 



In Fig. 122 let LM be the locus of [69], and let P be any 
point on LM. 



262 ANALYTIC GEOMETRY 

Then ,r=OK and y=PK. 

Let AX' be the line whose equation is 

, , A , AE D ' . 

[70] y=—^''^^F---B^ 

and let /" = QK, 

r n 1 ,/, ^ , AE D 

[71] then y ^ — -^^ -^ + ^^ — -;^- by §40. 

Then by subtracting [71] from [69], we get 

A 



[72] pQ__^_y" 



2B' 
A 



i'+i)' 



[73] Let V=VQ = - . p., 

let oi = -~, 

and let lY' be the line whose equation is 

C74] ^'=-i- 

E 

[75] Then gl^S = x — x' ^= x-\- —^. 

Then as in [28] of the first part 

[76] c^(^^k\x+~). 

Now if we take AX' for a new axis of X and lY' for the 
new axis of Y, then the coordinates of P will be 

[77] ^= ^Q and Y = PQ, 

[78] and X=k'<^x^~\ 

[79] hence ^ + — = ^. 



GENERAL EQUATION OF THE SECOND DEGREE 263 
Then [73] becomes 

[80] Y=-' ''' 



^3 X 2B'X' 
2B — 

k' 



[81] XY- ^ 



2B' 



k'A ... 

Since — -^^ is constant, [81] is the equation of an hyper- 

bola referred to its asNanptotes. by '§ 275. 

Therefore when C = o, [i] represents an hyperbola re- 
ferred to its asymptotes. 

SUMMARY 

C=o. 

Then when B' — AC <C o, and 

I St. CJ < o, [i] represents an ellipse; 

2nd. CJ > o, [i] represents an imaginary ellipse ; 

3rd. CJ = o, [i] represents two imaginary straight lines. 

When B' — ACy>o, and 

I St. CJ < o, [i] represents an hyperbola ; 

2nd. Cz/ > o, [i] represents the conjugate hyperbola; 

3rd. C/l = 0, [i] represents two straight lines. 

When B' — AC=o, and 

ist. C/l << o, [i] represents a parabola; 

2nd. C^ = 0, [i] reprasents two parallel straight lines. 

C=o. 

Then [i] represents an hyperbola. 

33i<2 . Corollary. — Every equation of the second degree contain- 
ing two variables only is the equation of a conic. 

Remark. — The lines represented by [43], [49] and [61] of 
§330 are limiting cases of the corresponding conies. 



NON-CONICS 

332. A Non-Conic. — Any plain locus which is not a conic 
we will call a non-conic. Among the non-conics are 

Higher Plane Loci, 

Spirals, 

The Logarithmic Curve, 

Trigonometric Loci. 

HIGHER PLANE LOCI 

333. A Higher Plane Locus. — A higher plane locus is a 
locus whose equation is of a higher degree than the second. 

The Lemniscate 

334. The Lemniscate. — If from the center of an equi- 
lateral h3^perbola a perpendicular be drawn to a tangent to the 
hyperbola, and the point of tangency be made to move along 
the hyperbola, the locus traced out by the intersection of the 
perpendicular and tangent is called the lemniscate. 




Let MAN be an equilateral hyperbola ; OT a perpendicular 
drawn from the center to the tangent which touches the hy- 
perbola at the point P, and let P move along the hyperbola. 



NON-CONICS 265 

Then the locus traced out by T, the intersection of the per- 
pendicular ajid the tangent, is the lemniscate. 



PROPOSITION 
335. The equation of the lemniscate is 

{x'^+fy=^a\x'^—f). 
Let X = OH and y = TH, 

a = OA. 

The equation of MAN is 

[i] x'—y'=a\ by §208. 

and since P is on the hyperbola 

[2] x"—y" = a\ by §40. 

Hence the equation of the tangent to the equilateral hyper- 
bola may be written 

x' a^ 
r^l r = — tX -. by §§ 221 and 207; 

■- y y 

Therefore the equation of the perpendicular OT is 

y' 

[4] JK = J X. by §§55 and 62. 

Since the point T is on both the perpendicular and the tan- 
gent, we may let the x andy in [3] and [4] stand for the co- 
ordinates of the point T. by § 40. 

Then [3] and [4] become simultaneous. Solving them we 
get 

[5] -^^ ^2 1 .,2 and y = — 



x'-{-y' -" x' + y'' 

Substituting these values into [2] and reducing, we get 
[6] {x'+yr = a'(x'-y). 



Q. E. D. 



266 ANALYTIC GEOMETRY 

336. Corollary. — The polar equation of the lem7iiscate is 

r' :=a''^ cos 2 6, 
in which r^iOT and 6 ^ ^AOT reckoned counter clockwise. 

This equation may be found by transforming [6] to polar 
coordinates. 

Scholium. — Let ^ = 0, then ;•= =h a. 

Let o<0<45°, 

then i>cos20>o. 

Hence • <2' cos2^ <<2'"^, 

and r^<Za'. 

Therefore r has two values numerically equal, less than a, 
and of opposite signs. 

Let 6 = 45°, then r =: o. 

Let 45° < < 135°, then r is imaginary. 

Let d = 135'', then r = o. 

Let 6 = 180°, then r = =b <a:. 

The curve therefore consists of two loops, one to the right 
and the other to the left of the origin, symmetrical with respect 
to the X axis, and reaching to the distance a from the origin. 

The Cissoid 

337. The Cissoid. — If on opposite sides of the center of a 
circle and at equal distances from it two ordinates be drawn 
to any diameter ; and if through the upper extremity of 
either ordinate and the extremity of the diameter farthest 
from it a line be drawn, and this line be made to revolve about 
the extremity of the diameter, then the locus of the point 
where this line cuts the second ordinate is the cissoid. 

Let QR and ST be two ordinates drawn to the diameter OA 
at equal distances from C and on opposite sides of it, and let 
OL be a line through S, the upper extremity of the ordinate 
ST, and O the extremity farthest from ST of the diameter OA. 



NON-CONICS 
(M 



267 




Fig. 124 

Let P be the point where OL cuts the second ordinate QR. 
Now let OIv revolve about the point O. 

Then the locus MPON traced out by P is the cissoid. 



PROPOSITION 
338. The equation of the cissoid is 



r = 



x-' 



2a — X 



Let X = OR, J^/ = PR and a = OC. 

[i] CR = CT, by construction. 

[2] hence OR = TA. 

[3] Again ST — OT.TA, by Geom. 60. 

[4] or ST = {2a — TA)TA. 



268 ANALYTIC GEOMETRY 
Hence by [2] 

[5] st'= (2^— 0R)0R, 

[6] or ST = i/x(2<a^ — x). 

Now OPR and OST are similar, by Geom. 51 

[7] hence OR : PR :: OT : ST, by Geom. 31 

[8] or X \ y w 2a — x : ST. 

Hence by [6] 



[9] X \ y \\ 2a — X : \/ x{^2a — jr). 



xy x{2a — x) 

[10] and v= — ^ z= rb \ 

2a — X \ 



x' 



2a — X 

3 



X 

fill Therefore y^ = . 

-^ 2a — X 

Q. E.D. 

Scholiuw.. — From [10] we see 

( 1 ) that the curve is symmetrical with respect to the X axis, 

(2) it passes through the origin, 

(3) it has the line x ■=i2a for an asymptote. 

339. Corollary, — The polar equation of the cissoid is 

r^=-2a tan 6 sin 6 . 

The Witch 

340. The Witch. — If to any diameter of a circle an ordi- 
nate be drawn and this ordinate be produced until the pro- 
duced ordinate is to the ordinate itself as the whole diameter is 
to either segment of the diameter, and then the ordinate be 
moved continually in the direction of this segment ; the locus 
of the extremity of the produced diameter is the witch. 

In the circle OLM let LK be an ordinate to the diameter 
OM. Let lyK be produced until 

[i] PK : LK :: OM : KM, 

and let PK move continually in the direction of the arrow. 




269 



Fig. 125 

Then NOPQ, the locus of the point P, is the witch. 
Let ^ = OK, jK = PK and 2a = OM. 



PROPOSITION 
341. The equati07i of tke witch is 

4.a~x 



y — 



2a X 



[2] 



OK : IvK :: LK : KM. 



by Geom. 60. 



[3] Hence LK = i/OK.KM = yx{2a —x) 

Therefore from [1] and [3] we get 
[4] y : -|/jr(2<2 — x) :: 2a : 2a — x. 



[5] Hence 



y 



^a X 



2a — X 

Q. K.D. 

342 . Co rolla ry . — 

(7) The witch is symmetrical tvith respect to the X axis ; 



270 



ANALYTIC GEOMETRY 



(-?) // lies wholly to the right of the Y axis ; 
(^) The line x ^=^ 2a is an asymptote to it. 

The Conchoid 

343. The Conchoid. — If while the center of a circle moves 
along a fixed straight line, one of its diameters always passes 
through a fixed point, then the locus of the extremities of the 
diameter is the conchoid. 




Figr. 126 

Let the center C of the circle PQP' move along the fixed 
line X'X, while the diameter PP' passes through the fixed 
point M ; then the locus SPT-LMN of the extremities P and 
P' of the diameter is the conchoid. 

Let P be any point on the locus and drawMts ordinate PK. 
LetJi: = OK and r = PK, 



PROPOSITION 

344. The equation of the conchoid is 

xY^{b'^-f){a + yy 

Draw MH 1| CK. 
Produce PK until it meets MH at H. 



NON-CONICS 271 

Since the triangles MPH and CPK are similar, byGeom. 25. 
[i] PK : CK :: PH : MH. by Geom. 31. 

[2] then PK ; VcP— PK :: PK + OM : OK, 

by Geom. 27 and 17. 



[3] or y : y^d'' — y \\y-\-a : x, 

[4] hence ;ry = (^^ — J'^)(^+J^')^ 



Q. K.D. 



From [4] we get the equation of the conchoid in another 
form . 

a-\-y 



[5] x^^-^Vb'-f. 

Scholium. — When ;i:=o, [5] becomes 

Hence the curve cuts the Y axis above and below the origin 
and b units from it. 

When y is numerically less than b, [5] shows that every 
real value of y, either positive or negative, gives two values 
of X numerically equal, but with opposite signs. 

Hence there is one branch of the curve above the X axis 
and another below it, and the curve is symmetrical with re- 
spect to the Y axis. 

When y = o, we find :r = ri= 00 , 

Hence the branches extend an indefinite distance to the 
right and left of the origin. 

Let b> a. 

Then when _y ^= — o., we have jr = o, 
and when j = — b, we have ;ir = o. 

Hence the curve cuts the Y axis below the origin at the 
distance — a and — b from it. 

When a <Cy< b, x has two values numerically equal, but 
with opposite signs. 

Hence there is a loop below the origin. 



272 ANALYTIC GEOMETRY 

If we take M for the pole and MY for the initial line and let 

r=MP and = YMP, 
we get the following : 

345- Corollary. — The polar equation of the conchoid is 

r = a. sec dz <^. 

The Limacon 

346. The Limacon of Pascal.- -If one of the two points 
in which a secant cuts a circle remains fixed while the other 
moves along the circumference of the circle, and if the length 
of the exterior segment of the secant produced through this 
latter point remains constant, then the locus of the extremity 
of this secant is the limagon of Pascal. 




Fig. 127 



In Fig. 127 let the point O remain fixed while Q moves 
along the circumference of the circle. OQQ'. lyCt the length 
of the exterior segment PQ remain constant. 

Then the locus of P is the limagon. 



NON-CONICS 273 

Let jr = OK and y = PK, 
^=OA *' >?=PQ. 

PROPOSITION 
347. 77z<? equation of the limago7i is 



[i] For OP = j+OQ = i/^^+y, byGeom. 26. 
[2] hence >? + ^.cosAOQ = |/^- 4-j/". 

by Geom. 55, Trig. 2. 
Now from the triangle OPK we get 

cos AOQ = — — ^ . 

t/ x'+y 

[3] Hence by [2] 

X ■ 



y x^-j-y 

[4] or ^-|/;t:^ -\- y^ -\- dx = x'"' -f" JI/^ 
[5] Hence (jr' +/— ^;r)=^ = ^^(jc'^+y). 



Q. K.D. 



By expanding [5] we get 

[6] y'-\-y\2x'—2dx — s') + x'\_{x — dy—s'l, = o. 

[7] Let ^^=j)/^ ^ = 2;r'^ — 2^^ — /, 

and e^iX^\_{x — d)' — /] . 

Then [6] becomes 
[8] 2j' + dv + c — o. 

By the theory of quadratics we know that v will be real when 

[9] d'' — 4.c > o, 

[10] that is, if (2^1;^ — 2dx — s^)^ — ^x'\_{x — d)^ — -f^] ^. Oj 



s' 



[11] hence if x -^ . 

— \d 

When X ^ d — s and x <^d-\- s, 



274 ANALYTIC GEOMETRY 

[12] c=:^x''\_{x — dY—s''']<o. 

For let x^^d — s-\-r^ 

then by the second inequalit}^ above 

d — s-\- r< ^-|-^. 

Hence r^ < 2 r^. 

Again let x^:^ d-\- s — r, 

then also r' < 2rs. 

But X — ^=^ — r, 

hence (^ — dy — ^^ = r"^ — 2rj ■< o, 

and x^[{x — d)^ — r] < o. 

That is, ^ is negative, and therefore one of the values of v 
in [8], say v^, is positive, and the other, say v^, is negative. 

Hence from [7] we get 

[13] • y=^i» 

[14] and y — — z/,. 

From [14] we see that two of the values of y in [8] will be 
inaginary, and from [13] that the other two will be real, equal 
and will have opposite signs. 

From [5] we see that 

ist, when ^ =: o and j}/=:o, the equation of the locus is 
satisfied. 

[15] Hence the limagon passes through the origin. 

2nd, when J/ = o, then x = dzL s. 

[16] Hence the limagon meets the X axis at two points. 

x:=:d-{-s and x = d — s. 

FIRST CASH 

Let s <C d. 
Then when x -< d — s, 

^ < o, 
and 2x'^ — 2dx — ^^ < — 2ds + /. 



NON-CONICS 275 

But s < d, 

hence 2X^ — dx — ^^ < o, 

[17] or ^ < o. 

Consequently in this case both the values of v in [8] are 
positive, and hence all four values of y in [6] are real and 
equal two and two. 

Hence we see by [11], [15], [16], that from 

s' 

X=z to X = O, 

and from xz:^o to x ■=. d — s, 

the limagon has two branches, both symmetrical with respect 
to the X axis. 

From [13] and [14] we see that from 

;r=^ — ^ to X ^= d -\- s, 

the limagon has but one branch, and it is symmetrical with 
respect to the X axis. 

Therefore when s <C d, the limacon will be of the form given 



in Fig. 127. 






SECOND CASK 




I^et s = d. 


In this case 


s' _ d 
\d~ 4 


and 


d — s = 0. 


Hence again when 


jf << d — s = 0. 




^ < 0. 



Therefore from to o, y will have four real values equal 

4 

two and two ; the points — s will be at the origin ; the loop will 
become the point O. From o to 2d, y will have two values 
equal and of opposite signs, and the limagon will take the 
form shown in the following figure. 



276 



ANALYTIC GEOMETRY 

Y 




Fig-. 128 

The lima^on is then called the cardioid. 
Since in this case s = d, [5] becomes 

[17] (^^ +f - dxY^d^x' +y ), 

which is the equation of the cardioid. 



THIRD CASE 

lyCt d <^s <Z 2^. 
Then d — s is negative. 

Also when x <^ d— s, 

we have <5 < o. 



Therefore from 



^d 



to d — s the J/ in [5] has four real 



values equal two and two with opposite signs. From^ — s to 
d -\- s it has two real and equal values with opposite signs. 

Since jr = o andjK = o satisfy [5], the origin is an isolated 
point on the limagon. Hence in this case the limagon takes 
the form shown in the following figure. 




277 



Fig, 129 



Then 



/^d 



FOURTH CASK 
I^et S =: 2d. 

= — d and d — s =^ — d. 



Now when a- = — d in [8] , ^ = o and r = o and the four 
values of y in [6] become o. 

From — d to 3d, j/ has two real and equal values with oppo- 
site signs. Therefore the limagon takes the form of Fig. 130. 




Fig. 130 



278 ANALYTIC GEOMETRY 

FIFTH CASE 
Let ^ > 2d. 

2 

Then the values of y in [6] will be imaginary from ^ 

4^ 

to ^ — s, but from d — ^ to d-\- s, y will have two real and 

equal values with opposite signs. Therefore the limagon will 

take the form shown in Fig. 131. 




336. Corollary. — The polar equation of the limagon is 

r=^s-\-dcosd. 
In Fig. 127 let O be the pole and POX the vectorial angle. 
Iyet(9=POX and r=OP. 

Now AQO is a right angle, by Geom. 55. 

[i] hence OQ = OA cos := d cos 0. by Trig. 2. 

[2] But ' ^ = PQ + OQ, 

[3] hence r = .y -f- ^cos ^. 

Q. K. D. 



NON-CONICS 279 

Scholium. — If we take the point P', then 

[4] OQ'= OA.cosAOQ'=— OAcos AOP', by Trig. 23. 

[5] or 0Q'=— ^cos6>. 

[6] But r=OP' and 5 = Q'P', 

[7] hence r z^ s — OQ'. 

Substituting the value of OQ' given in [5] into [7], we get 

[8] r=s-\-dcosO. 

If we take the point P", then 

[9] _oP- = — OQ" + Q"P", 

[10] or OP'' = OQ" — Q"P". 

[11] Hence rz=^s-\-d(ios>6. 

SPIRALS 

346. A Spiral. — A spiral is the locus traced out by a point 
revolving in a plane about a fixed point and receding from it 
according to some fixed law. 

Logfarithmic Spiral 

347. The Logarithmic Spiral. — The logarithmic spiral is 
the locus of a point revolving about a fixed point in such a 
way that the logarithm of its radius vector is always equal to 
a constant multiplied by the number of radians in its vectorial 
angle. 



28o ANALYTIC GEOMETRY 

PROPOSITION 

348. The equation of the logarithmic spiral is 

in Q 




Fig. 132 

In Fig. 132 let P move about the fixed point O in such 
a way that the logarithm of OP is always equal to a constant 
multiplied by the number of radians in ZlPOD. Then will 
ABCDK, the locus traced out by P, be the logarithmic spiral. 

Let O be the pole and OX the initial line. 
Ivetr=OP, 

= the number of radians in z^POD, 
and m = any constant. 

By the definition of the spiral we have 

[i] log r^^md. 

Then by the definition of a logarithm we have 

[2] r = «^^®. 

Q. E. D. 

PROBLEM 

349. To construct the logarithmic spiral whose equation is 

[i] r=2d. 

The number of degrees in one radian is 57°. 3. by Trig. 27. 
Therefore by [i] we may make the following table. 







NON-CONICS 








TABLE 




Number of degrees in 


e- 


e. 


Length of r. 


O 







I 


57"-3 
ii4".6 

57-3 
— 114. 6 




I 

2 

— I 

2 


2 
4 

•5 

•25 


— 00 




3 






281 



Since by the table r= i when =0, tlie point whose coor- 
dinates are r= i, =0, is the point D in Fig. 132 at the dis- 
tance of a unit from the pole. 

Since by the table when the number of degrees in 6^ = 57°. 3, 
r=: 2, we may locate a second point P on the spiral by laying 
off Z^ POX = 57°. 3 and making OP == 2. Similarly we may 
locate the point K. 

Since by the table r = 0.5 when 6 =: — 57°-3> we may 
locate the point C in the figure b}^ laying off an angle of 57^.3 
measured from OX clockwise, and making OC = 0.5. Simi- 
larly we may locate the point B. 

In this way we may locate any number of points and draw" 
the spiral. 

350. Corollary i. — Since when =z o, r =^ i, we see that any 
logarithtnic spiral cuts the initial lijie at a unifs distance from 
the pole. 

Corollary 2. — Since when = — co , r = o, the spiral makes an 
infinite number of revolutions within the circle whose radius is i. 

Corollary j. — Si?ice when ^ = 00 , r = co , the spiralmakes an 
i7ifinite number of revolutions outside of the circle whose radius 
is I . 

The Spiral of Archimedes 

351. The Spiral of Archimedes. — The spiral of archimedcs 
is the locus traced out by a point moving about a fixed point 
in such a way that the ratio of its radius vector to its vector- 
ial angle is constant. 



282 



ANALYTIC GEOMETRY 




Fig". ^33 

Let the point P revolve about the fixed point O in such a 

OP 
way that ^^^^ is constant ; then the locus OMPQ is the spiral 

of archimedes. 



PROPOSITION 

352. The equation of the spiral of archimedes is 

r= c.d. 

Let r = OP and 6 = the number of radians in POX and 
c = a constant. 

Since by definition 

[2] r=^c.e. 

Q. K.D. 

353. Corollary.— 

(i) Since when B^^o^r^=^o, the spiral passes through the pole ; 
{2) Since when = 00 , r=: 00 , the spiral makes an infinite 
number of turns about the pole. 

The Hyperbolic Spiral 

354. The Hyperbolic Spiral, or the reciprocal of the spiral 
of Archimedes. — The hyperbolic spiral is the locus traced out 



NON-CONICS 



283 



by a point revolving in a plane about a fixed point in such a 
way that the product of its radius vector and the number of 
radians in its vectorial angle is constant. 







Y 


^ 






/ 


/ 


\ 


\ 


/ 


\ 0. 


( \ 




^ 




Q^ 


IS 

H 
Y' 




D 



Fig-- 134 



In Fig. 134 let O be a fixed point and OX a fixed straight 
line. IvCt P revolve about O in such a way that OP multi- 
plied by the number of radians in ^ POD is constant ; then 
will PQRS, the locus traced out by P, be the hyperbolic spiral. 



PROPOSITION 

355. The equation of the hyperbolic spiral is 

rd=c. 

Let r = OP, = the number of radians in POX, and c^=^ 
a constant. 

Then by definition we get 

[i] rd = c. 

356. Corollary. — 

(z) The spiral makes an infinite number of revolutions about 
the pole before reaching it ; 

c 
{2) Si?zce r = -^, there is no point on the spiral whose vec- 
torial angle is zero. 



284 ANALYTIC GEOMETRY 

357. Corollary 2. — The constant c is the circuynference of a 
circle whose radius is equal to the length of the radius vector at 
the end of the first revolution. 

For at the end of the first revolution 

[2] Q = 271. 

lyet r' = r at the end of the first revolution. 

Then by [i] 

[3] 2nr'=c, 

and 27tr' = the circumference of a circle whose radius = r\ 

by Geom. 29. 

358. Corollary 3. — The arc of a circle between ajiy point on the 
spiral arid the ijiitial line is equal to the circumference of the cir- 
cle whose radices is the lerigth of r at the end of the first revolu- 
tio7i . 



^4] For 


r =^' 


by Trig. 28. 


[5] or 


arc PD = rQ. 




Hence by [i] 






[6] 


arc PD = c. 





Therefore by §357 the arc PD is equal to the circumfer- 
ence of the circle whose radius is equal to the length of r at 
the end of the first rev^olution. 



The Parabolic Spiral 

359. The Parabolic Spiral. — The parabolic spiral is the 
locus traced out by a point revolving in a plane about a fixed 
point in such a way that the ratio of the square of its radius 
vector to the number of radians in its vectorial angle is con- 
stant. 




285 



Fig. 135 



PROPOSITION 



360. The equation of the parabolic spiral is 



e 



'= c. 



For the equation follows at once from the definition. 



The Lituus 

361. The Lituus. — The litzms is the locus traced out by 
a point revolving in a plane about a fixed point in such a way 
that the product of the square of its radius vector and the num- 
ber of radians in its vectorial angle is constant. 

Y 




Fig. 136 



286 



ANALYTIC GEOMETRY 
PROPOSITION 



362. The equation of the liticus is 



r'd = c. 



The equation follows at once from the definition. 

363. Corollary, — The initial li7ie is an asymptote to the lituus. 

THE LOGARITHMIC CURVE 

364. The Logarithmic Curve. — The logaidthmic curve is 
the locus of a point moving in a plane in such a way that its 
abscissa is always equal to the logarithm of its ordinate. 




Fig- 137- 

Let P be any point in the plane YOX. 

Let P move in the plane YOX so that OK is always equal 
to the logarithm of PK. Then RPQ will be the logarithmic 
curve. 



PROPOSITION 

365, The equation of the logarithmic curve is 

y ^^ a^ . 
Let jr = OK and 7 = PK. 



NON-CONICS 

Then by the definition of the curve 
[i] x = \Qgy, 

hence, by the definition of a logarithm, 
[2] y = a''. 



287 



Q. E. D. 



366. Corollary i. — The whole of the logarithmic curve lies 
above the X axis. 

For since negative numbers have no logarithms, y can never 
be negative in [2] . 

367. Corollary 2. — Every logarithmic curve must cut the Y 
axis at a point one unit above the origin. 

For when jf =r o, 

[3] 7=^^° = !, 

whatever may be the value of the base a. 

368. Corollary j. — The X axis is an asymptote to the curve. 

For when x = — co . 



y 



a' 



= o. 



a' 



TRIGONOMETRICAL LOCI 

369. A Trigonometrical Locus. — A trigonometrical locus is 
a locus, one of whose rectangular coordinates is a trigonomet- 
rical function . 

The Cycloid 

370. The Cycloid. — If a circle roll upon a fixed straight 
line, the locus traced out by a given point on the circumfer- 
ence of the circle is called a cycloid. 




288 ANALYTIC GEOMETRY 

lyCt X'X be the fixed straight line, and APBD a circle roll- 
ing upon that straight line in the direction of OT. Let P be 
a given point on the circumference of the circle. 

Then the locus OPQRS, traced out by the point P, is a 
cycloid. 

PROPOSITION 
371. The equation of the cycloid is 

y 

X z=:.r vers—^ \/ 2ry — r^. 

r \ ^ - 

In Fig. 138 let 

* X = OK and y = PK, 

;. = cp " = z:pci.. 

[i] Now OK = OA — KA, 

and, since the circle rolls in the direction OT, 
[2] OA = the arc A P. 

[3] But the arc AP = rd, by Trig. 28. 

[4] hence OA = rO. 

[5] Again KA = Ply, by Geom. 17. 

[6] but PL = r sin(9, by Trig. i. 

[7] Hence KA=rrsin^. 

Substituting the values of OA and KiV found in [4] and 
[7] into [i] , we get 

[8] OK = r<9 — r sin Q. 

[9] Therefore x = rd — r sin 0. 

[10] AgainPK= AL = CA — CL. by Geom. 17. 

[11] Now . CL = ^^ cos ^, by Trig. 2. 

[12] hence PK = CA — rcos 0. 

[13] PK —y and CA = r. by Geom. 18. 



NON-CONICS 289 

Substituting the values of CA and PK found in [13] into 
[12], we get 

[14] y = ^ — rcos 6. 

Multiplying both, sides of [14] by 2r, we get 

[15] 2?y = 27^ — 2r' cos Q . 

Squaring both sides of [14], w^e get 

[16] y'^ =^ r' — 2r' cos^ + ^'' cos"0. 

Subtracting [16] from [15], we get 

[i 7] 2ry — j/^ = r" — r^ cos ^Q =^7^ ( i — cos~ Q) . 

Now since i — cos^ d = sin^ Q^ by Trig. 5. 

equation [17] becomes 

[18] 2ry — JK^ =^ r^sin^ Q , 



[19] and -\/2ry — :j/^ = r sin 0. 
Hence [9] becomes 



[20] X ^=^ rd — y' 2ry — j/". 

CI/ 

[21] Now cos = , by Trig. 2. 



r22l hence cos = =^, 

r 

[23] or ^ = cos-i -. by Trig. 30. 

Hence [20] becomes 

[24] jr = r cos— ^ — |/ 2ry — y\ 

[25! cos— '^ ~ = vers—'- ( i — ) = vers-^— . 

r \ r .' r 

by Trig. 29. 
Hence [24] may be written 

[26] X =^r vers-i-=^- ]/ 2ry — y- . 

Q. E. D. 



290 



ANALYTIC GEOMETRY 



372. Coi'ollary. — When the highest point of the rolling circle 
is taken as the origin, and the diameter perpeiidicular to the line 
upon which the circle rolls as the X axis, the equation of the 
cycloid is 



yz^r vers^ (- -[/ 2rx 



X' 




Fig. 139 

In Fig. 139 let AT be the fixed straight line upon which 
the circle ABOD rolls in the direction of the arrow. Let P 
be any given point on the circle. 

Let x^OK and y = PK, 



[i] Then 


PK = 


^PJ + JK. 




[2] Now 


PN = 


LM, 


by Geom. 57 


I3] hence 


PJ- 


LK. 


by Geom. 59 


'j\\ Again 


JK = 


: HA, 


by Geom. 17 


hence [i] becomes 








[5] 


PK = 


-- HA+LK. 





NON-CONICS 

[6] Again AR == the arc AI^O, 

[7] and HR = the arc HP, 

[8] hence HA = the arc PQ. 

[9] But PQN = IvOM, 

[10] hence the arcPQ = the arc LO, 

[11] hence by [8] HA == the arc LO. 
[12] lyK = r sin 0. 



291 



by Geom. 58. 
by Geom. 59. 



Hence by [11] and [12], [5] becomes 

[13] PK --= the arc LO + r sin 6. 

[14] Hence y =^ rS -{- r sin 6. by Trig. 28. 

[15] Also ;r = OK = CO — CK = r— rcos0. by Trig. 2. 

Now from [14] and [15], as in the demonstration of the 
proposition, we may get 



[16] 



y ^ r yers 



+ |/^ 2rx — x'^ 



373. Other trigonometrical loci are the following : 
The curve of sines, see page 17. 
The curve of tangents, see page 18. 

The curve of secants, see page 19. 



SOLID ANALYTIC GEOMETRY 



CHAPTER I 

Points and Directions in Space 

374. Ihe posiiicn of a point in space may he indicated by 
means of its distances from each of three well kiiown, fixed ^ in- 
tersecting planes which are pcrpe7idicular to each other. 

Thus, in Fig. 140, let AB, CD and EF be three well known, 
fixed, intersecting planes perpendicular to each other. I^et P 
be any point in space, and draw 

PS _L to the plane AB, 
• PQ J_ " " " EF, 
PM J_ " " " CD. 

Then the position of the point P is indicated by giving the 
lines PS, PQ, and PM. 

375. ^The Coordinate Planes. — The three well known, 
fixed, intersecting planes are called the coordinate planes. 

The plane AB is called the ZX plane. 
( ( * s C\V) " " " Z Y ' ' 

( ( ( ( TTF " ' ' " X Y ' ' 

376. The Origin, — The point in which the coordinate planes 
intersect each other is called the origin. 

'^'j']. The Coordinate Axes. — The lines in which the coor- 
dinate planes intersect each other are called the coordinate 
axes. 



294 



ANALYTIC GEOMETRY 



The line in which the plane ZX intersects the plane XY is 
called the X axis. 

The line in which the plane ZY intersects the plane XY is 
called the Y axis. 

The line in which the plane ZX intersects the plane ZY is 
called the Z axis. 




Fig. 140 

378. Corollary. — The axes are perpendicular to each other. 

For the plane ZX _L the plane ZY, by construction 

and " " XY_L " " ZY, by construction 

Hence the line XX' _L " " ZY, by Geom. 44 

Hence XX' must be _L to YY' and ZZ'. by Geom. 33 

Similarly it may be proved that ZZ' is _L to YY' and XX' 

and that YY' is J_ to ZZ' and XX'. 

379. The Coordinates of a Point. — The three distances of 
a point from the coordinate planes are called \\\^ coordinates of 
the point. 

The coordinates of P are PM, PS and PQ, and they are 
respectively parallel to the axes XX', YY' and ZZ'. 

For PM is _L the plane ZY, by construction. 

and it w^as shown in § 378 that XX' is _L the plane ZY. 



POINTS AND DIRECTIONS IN SPACE 295 

Hence PM || XX'. by Geom. 35. 

Similarly it may be proved that 

PS II YY' and PQ || ZZ'. 
The coordinate || to the X axis is called the x coordinate. 

< ( ( ( II I i ( ( -VT" * < n 4 1 a ^. (1 

<( (( II ((((^ k( (( (( <( _ (( 

Let X = the x coordinate. 

2=. " ^ 

Pass a plane through PS and PQ ; another through PS and 
PM, and another through PM and PQ. 

PS is J_ the ZX plane. by construction. 
Hence plane PQRS *' J_ " ZX " by Geom. 43. 

Also PQ " _L " XY " by construction. 

Hence " PQRS " _L " XY '' by Geom. 43. 

Hence since the plane ZX is J_ to theplane PQRS, 
and " " " XY " J_ " " " PQRS, 

their intersection XX' '' _L " " " PQRS. 

by Geom. 44. 
Hence OR J_ SR. by Geom. 33. 

Now OL _L OR, by § 378- 

and we have just proved that 

SR ±. OR. 
Hence ' OL || SR. by Geom. 46. 

Similarly it may be proved that 

SL 11 OR. 
Hence OLSR is a parallelogram. 

It may also be proved that OLMN, MNQP, PQRS, ONQR 
and MLSP are all parallelograms. 

Hence PM = QN -= OR = :*:, 

PS = ML = ON = y, 
and PQ = MN=OL = ^. by Geom. 47. 



296 ANALYTIC GEOMETRY 

Remark. — The coordinate planes need not be taken per- 
pendicular to each other as above. When they are so taken 
the system of coordinates is then called a rectangular sys- 
tem, otherwise it is called an oblique system. 



PROPOSITION I 

380. If p be the distajice from the origin to any point and x, 
jy, z be the coordinates of that pointy then 

lf = x'+y' + 2\ 

In Fig. 140 let P be any point in space. 
Let p = OP be its distance from the origin. 
-^ = 0R, j = RQ andz = PQ. 
Now PQ _L OQ. by Geom. 33. 

[i] Hence ff = OQ + 2'^ by Geom. 26. 

[2] But QQ = X- +y\ by Geom. 26. 

[3] Hence p^ =^ x- -\- y- -\- 2'^ . 

Q. E. D. 

381. The Direction Cosines of a line drawn through the 
origin. — The cosines of the three angles which any line pass- 
ing through the origin makes with the axes are called the 
direction cosines of the line. 

In Fig. 140 let oi=,Z. POX, 

^^^POY, 

Then cos «', cos ^, and cos y are the direction cosines of 
the line OP. 

PROPOSITION II 

382. If X, y ,z be the coordinates of any point in space, p the 
length of the line drawn from the origin to that pointy afidcos a, 
cos /^, cos y the direction cosines of this line^ 

then X =^ p cos a, 

y = p cos /?, 
z = p cos y. 



POINTS AND DIRECTIONS IN SPACE 



297 



For in Fig. 140 OR _L PR. b}^ Geom. 33. 

[i] Hence OR = OP cos POX, by Trig. 2. 

[2] or X = p cos a. 

Similarly for j/ and 2. 

PROPOSITION III 

383. If cos (X, cos /?, cos y be the direction cosines of any line 
drawn through the origin^ then 

cos'^ oc -\- COS' ft -f- COS' y =^ I. 
For X' = p' COS" a^ 

y = p- cos^ ft, 

z' =■ p^ cos^ y. by § 382. 

[i] Hence ^' H-J^' ~h -s'^ = /^^(cos^ <a^4~ cos"' /? -(- cos^>^). 

[2] But x-^y'^-^-z''— p\ by §380. 

[3] Hence p^ = /o'(cos'^ a+ cos^ y^ + cos"' 7) , 

[4] and cos^ oc -\- cos"^ ft + cos' y ■:=. \. 

O. E. D. 



PROPOSITION IV 

384 . If x\ y' , z' be the coordin ates of a ny poin t i7i space ; x" ,y" , 
z" the coordinates of aity other point in space ; a7id D the length 
of the line joining these points, then 

D'= {x" —x'y+ {y" —yy+ {z" —z'y. 




Fig. 141 



298 ANALYTIC GEOMETRY 

In Fig. 141 let P' and P" be the two points. 
Z>=P'P", 

x' = OL, y = IvK, z' = P'K, 
jr"=OM, y'=MN, y'=P"N. 

We are to prove that 

D''— {^x" — x'Y -^ {y'—/y+ {2"— 2')' 
As in §380, it ma}^ be shown that 



[i] W" — P'R + RQ + P"Q. 

[2] But V'V. — lL] — l^U = x"—x\ byGeom. 17. 

[3] RQ=: JN = KH=y'~y, byGeom. 17. 

[4] and P"Q = -?" — ^'. 

Substitutino^ these values into [i], we get 

[5] z^•-'=(x"-^')^+(y'-y)^+(y'-y)^ 

Q. E. D. 

385. The Angle Between Two Lines in Space. — By the 

angle between two lines iji space ^ which do not intersect, we 
mean the angle between either of them, and a line drawn 
through any point on it parallel to the other. 

386. The Direction Angles Of Any Line in Space. — By 
the direction angles of any line in space we mean the angles 
w^hich a parallel to the given line drawn through the origin 
makes with the axes. 

387. The Direction Cosines of Any Line in Space. — By 

the di7'ectio7i cosines of any line in space we mean the direction 
cosines of a parallel to this line drawn through the origin. 

PROPOSITION V 

388. If x\ y\ z\ and x^\ y", 2" be the coordinates of any two 
points in space, p the le?igth of the line joining them, a7id cos ct, 
cos fi, cos y the direction cosines of that lijie, then 

x" — x' = p cos oc, 
y" — y =^ p cos p, 
2" — 2' =: p COS y. 



POINTS AND DIRECTIONS IN SPACE 299 

In Fi^. 141 let OP'" be drawn through the origin |! to P'P". 
I.et«=P'"OX, /i = P'"OY, K = P'"OZ. 
[i] Then P'R = P'P" cos P'T'R, by Trig. 2. 

[2] or x" — x' = p cosV"?'V.. 

[3] Similarly /' —y' = p cos P"P'V, 
[4] and 2-" — z' = p cos P"P'T. 

[5] But P"P'R = P'"OX = a, 

P'T'V -= P'"OY = /3, 
and p'/p'T = P'"OZ = ;/. by Geom. 11. 

Hence from [2], [3] and [4], we get 
x" — x' zrz p cos a, 
y — y ^= P cos y5, 

Q. K.D. 



2" — 2' z=^ p cos y 



PROPOSITION VI 



389. If cos oc^ cos P, cos y be the direction cosines of any line 
in space, and /, w, 71 be any three quantities proportioiial to them, 
then 



cos OL = 



-j//^ + ni^ + ^^ 



m 

cos p = 



COS y 



^r + 77i' + n' 
n 

-|//~ -\- vi^ -\- n^ 



I m n 

\i\ For =• -,,== . by hypothesis. 

^ -■ cos c^ cos p cos y 

_ / 

[2] Now let r^ . 

^ -* cos a 

Then from [i] we also get 

m 71 ' 

[3] ^ ^= 7? ^^^ ^ ^^ 



cos fi cos ;r* 



300 ANAL YTIC GEO ME TR Y 

From [2] and [3] we get 

[4] I z=z r cos <a', in = r cos y^, 71 ^=- r cos y. 

Then by squaring and adding the corresponding mem- 
bers of [4] , we get 

[5] /" + m^ + 71^ = r' (cos''' OL -f- cos* /5 + cos" x) . 

[6] But cos^ «' + cos~ /^ + cos^ 7 = I , by § 383. 

[7 J hence P -\- in~ -\- n' = r\ 



[8] and r — ^/ /'' -j- //i' -j- 71'. 

From [4] we get 

_/_ 
r ' 

m 

r ' 
n 



[9] cos OL = 

[10] cos/5 = 

[11] cos y = 



Now substituting the value of r found in [8] into [9] , [10] 
and [11] , we get 

/ 



12] 


cos Oi = 




^l^ + m' + n" 


[13] 


_ 1^1 


LUb Z-' - — 




l//-^ + nt" + 7i' 


[14] 


n 





^l' + m' + n' 

Q. K. D. 

390. Corollary. — We can determine the directio7i of any line 
by means of any three numbers proportional to the direction 
cosines of that line. 

391. Directors of a Line. — Any three numbers propor- 
tional to the direction cosines of a line are called directors of 
that line. 

392. Corollary. — The direction cosines of a line are directors of 
that line. 



POINTS AND DIRECTIONS IN SPACE 301 

For in [4], § 389, r may be any number whatever. It may 
therefore be i. In that case we have 

/ zr= cos a^ m = cos y^, n 1= cos y. 



PROPOSITION VII 

393. If the directors of any line are proportional to the directors 
of a second line ^ the two lines aj^e parallel. 

For let /, vt and n be the directors of the first line., 
and let /', ??^' and n^ be the directors of the second line. 

by hypothesis. 



[i] Then 




I' _ w'__ ?l' 

I in n ' 




[2] Let 




V 




Then from [i] 


we 


get 




•[3] 


r : 


= — and r := 
in 


n 
n 



Hence from [2] and [3], w^e get 

[4] /' = r/, in^ = rm^ and 7i^ = rn. 

Now let cos ^, cos /?, cos y be the direction cosines of 
the first line, and let cos o(.\ cos (i\ cos y be the direction 
cosines of the second line. Then by § 389, [12] , [13] and [14] 
and by [4] above, w^e get 

/' rl 

[5] cos ^' = 



^r + in!' + n'-' 1/ r'l' + r'nt' + r'n/ 
I 



cos a. 



yU- + m' + n' 

|_6] Similarly cos /^' = cos /?, 

[7] and cos y = cos y . 

Now since the direction cosines of the two lines are equal, 
the lines are parallel. 

Q. E. D. 



302 



ANALYTIC GEOMETRY 
PROPOSITION VIII 



394. If o^, /5, y ai^e the direction angles of any line ; a\ j5\ v' 
the direction angles of any other line, and Vis the angle betiveeu 
the two lines, the7i 

V = cos a cos a' + cos fS cos ft' -\- cos y cos y'. 




Fig- 142 

Let x^' = OH, y" = HK, 2" = PK, 
x' = OH' y = H'K', y = P'K', 

^ = pox, /i = POY, r = POz, 

a' = P'OX, /i' = P'OY, f = P'OZ, 

p" = OP and p' = OP'. 

[i] Then 
x" = p"cosa, y =1 p" cos /?, and 2:"=:p''cosy, by § 3S2. 

[2] and 

x' = p' cos «', y = p' cos (3\ and 2' ■= p' cos y' , by § 382, 

[3] and D'={x" — x'y+{y"—y'Y+{2" — 2'y. 

by § 384. 
Hence by [i] and [2] we get 

[4] D^= ip" QOS>a — p' cos a'Y+ ( p" cos /3 — p' cos /3' )' 
+ ip" cos y — p' cos y')\ 

[5] Hence Z>- = p"^ (cos* <a' + cos^/?-|- cos~ ;k) 
+ p'- (cos^ a' + cos^ /3' + cos'V') 
— 2p"p' (cos «' cos a' -\- cos /3 COS y5' -|- cos y cos x') • 



POINTS AND DIRECTIONS IN SPACE 303 

[6] Hence 

E>-=p"--\-p'- — 2 p'p" {cos a cos <a:'-l-cos /^cosy^' + cos y cos;k'). 

[7] But i^^ =p""' — 2p'p"cos F+p'-. by Trig. 25. 

[8] Hence p"- + p" — 2p'p" cos F = 
p"- -|- p'- — 2p'p" (cos a cos «'' -|-cos ft cos /^' -f- cos y cos x')- 

[9] Hencecos F=:cos<^cos<a:' + cos/?cos/5'-|- cos/cos x'. 

Q. E. D. 

395. Corollary i — If cos a, cos /?, cos y, and cos a\ cos /^', 
cos y' be the direction cosines of two lines, then in order that the 
two lines be perpejidicidar to each other we must have 

[10] cos a cos OL^ -j- cos (3 cos (3' -j- cos y cos y' =:^ o. 

For in order that the two lines shall be perpendicular to each 
other, we must have in [9], § 394 

cos K= o. by Trig. 19. 

396. Corollary 2 . — If /, m, n, and /', ni\ n' be directors of 
two lines, then in order that the two lines shall be perpendicu- 
lar to each other, we must have 

IV -j- mm^ -\- 7i7i' = o. 
For as in §389, [9], [10] and [11] 

— z= COS oc, — = COS /?, and — = cos y ; 
r r r 

V 171 n 

and — r = cos oc\ — ;- = cos /5'', and — r ^ cos y'. 

r r r . 

Substituting these values of the direction cosines into 
[10], § 395» we get . 

^ ^ //' + mm' + nn' 
[11] , = o. 

[12] Hence //' + min' + nn' 1= o. 



304 ANALYTIC GEOMETRY 

PROPOSITION IX 

397. If a, b, c, and a\ b\ c' be the direction cosines of any two 
lines ^ and V be the aiigle between them, then 

sin'' V— {ab' — a'b)-+ {be' — b'cy+ {ca' — c'a)-. 
[1] For sin" F= i — cos'' V. by Trig. 5. 

[2] But i—a' + b'+r. by §383. 

[3] Hence sin'^ V =: a'- -{- b^ -{- c' — cos^ V. 

But b}^ § 394 

[4] COS' [' = 

a'a'' + b'b'' + rd' + 2aa'bb' + 2bb'cc' + 2cdaa'. 

[5] Hence sin' F = 
a' + ^'-p r"' — a' a'' — b'^'b'' — cd' — 2aa'bb' — 2bb'cd — 2cdaa\ 

[6] Hence sin' V — 
a\i—a")+b\i — b"') + c-{i~c") — 2aa'bb'—2bb'cd—2ccaa\ 

But by § 383 

[7] i—a" = b''' + d\ i—b''^a"+d' 

and i—c" — a"-\-b'\ 

[8] Hence sin' V = 

a\b" + d'-) -h b\a" + ^') + c\a" + b''') 
— 2aa'bb' — 2bb'cd — 2cdaa\ 

[9] or sin' F= a"V'' — 2aa'bb' -|- a'^'b~ 

+ b'd'— 2bb'cd + b''r 
a c — 2aa cc "^ a 'c . 

[10] Hence sin' F = 

{aU — a'by^ {bd — b'cf-\- {ca' — day. 

Q. E. D. 



CHAPTER II 

Projections 

398. To project any line upon a second line, draw from 
each extreme of the first line a perpendicular to the second. 
The segment of the second line between these perpendiculars 
is called the projection of the first line upon the second. 




Fig-. 143 



To project the line extending from A to B upon the 
line X'X, we draw a perpendicular from A and another from 
B to the line X'X. Suppose these perpendiculars meet X'X 
at the points K and L,. Then the line extending from K to 
L, is the projection of AB on X'X. 

399. If we take the direction X'X as the positive direction, 
then Kly will be positive, and lyK will be negative. 

400. lyct the projection of any line AB upon the X axis be 
represented by ABx , its projection upon the Y axis by ABy , 
and its projection upon the Z axis by ABz . 

401. The Projection of a Broken Line. — The projection 
of a broken Ime is the algebraic sum of the projections of the 
segments of that line. 



3o6 



ANALYTIC GEOMETRY 




Fig. 144 



Let the direction X'X be the positive direction. 
The projection of ABCD on X'X is 

KIv — ML + MN = KN. 



PROPOSITION I 

402. The projection of any closed contour upon a straight line 
is zero. 




r^ 



M N 

Fig. 145 



■»x 



Let X'X be any straight line. Let ABCDKA represent any 
closed contour. 

We are to prove that 



ABCDEAx = o. 
According to § 400 we will let 

AB, = KL, 
BCx =LM, 
CDx = MN, 
DKx = NO, 
EAx = OK. 



PROJECTIONS 



307 



[2] 



ABCDx=KIv + I.M + MN=: KN. by § 401. 
~ :NO + OK=NK=— KN. 

by § 401 and § 399. 



DEAx 
Then by adding [i] and [2], we get 



ABCDEAx = o. 



PROPOSITION II 



Q. K.D. 



403. When any two broken lines or a straight line and a broken 
line have the same extremities their projections on the same li?ie 
are equal. 









B 






c- 








A 


( 






F 








\ 


\p 


^^ 
























^ V 


iP/Z'W 


\ 


: 


L 1 


' ( 


) n 


/I 


\ 


>A 



Fig. 146 

Let ABC and ADEFC be two broken lines having the same 
extremities A and C. 

Let X'X be any straight line upon which ABC and ADEFC 
are projected. 

We are to prove that 



ABCx = ADEFCx. 

Now since the two broken lines form a closed contour, we 
have 



[i] KL + EM + MN + NO + OP + PK = o. by § 402. 
[2] Hence KE + EM = — MN— NO — OP — PK, 
[3] and KE + EM = KP + PO + ON + NM. 

by § 399. 

[4] or ABCx = ADEFCx . by § 401 . 

Q. E. D 



3o8 



ANAL YTIC GEO ME TR Y 



PROPOSITION III 

404. The projection of any straight line tipon a second straight 
line is equal to the length of the first multiplied by the cosine of 
the a7igle between them. 




Fig. 147 

Let AB be any straight line and X'X the line upon which 
we wish to project it. 

We are to prove that ABx is equal to AB times the cosine 
of the angle between AB and X'X. 

Through A and B pass planes _L to X'X at the points K and D. 
Draw AC || X'X and piercing the plane MN at C. 

Draw BD and BC. 

Since by construction ED is _L to the planes, MN and OP, 
it must be _L to BD and AH. by Geom. 33 

Hence ED is the projection of AB upon X'X. by § 398 



The plane MN is |1 to the plane OP. 



[i] Hence 



AC= ED. 
AC II X'X. 
AC_Lthe plane MN, 

AC J_BC. 
AC = AB cos BAC. 



Hence 

and 
[2] Then 

Hence by [i] we get 

[3] ED = AB cos BAC. 

Now AC II X'X, 

hence BAC is the Z. between AB and X'X. 



by Geom. 38 
by Geom. 40 
by construction 
by Geom. 36 
by Geom. 33 
by Trig. 2 



by construction. 

by § 385. 



PROJECTIONS 



309 



Therefore by [3], KD, the projection of AB on X'X, is 
equal to AB multiplied by the cosine of the angle between AB 
and X'X. 

Q. E.D. 

PROPOSITION IV 

405. If two straight lines be parallel, their projections 2ipon the 
same pi a 71 e are pa rallel. 




Fia:. 148 

Let PB and KD be two parallel straight lines. 

Let BQ and DM be their projections upon the plane ZX. 

We are to prove that 

BQ II DM. 

Since Q is the projection of the point P on the plane ZX, 
PQ is _L the plane ZX. byGeom.34. 

Hence the plane PBQ is _L the plane ZX. byGeom.43, 
Similarly the plane KDM is J_ the plane ZX. 
Let R be any point on the line PB and draw 
RH and PK _L DK. 
[i] Then RH || PK. by Geom.46. 



3IO ANALYTIC GEOMETRY 

Let S be the projection of the point R upon the plane ZX. 
[2] Then RS || PQ. by Geom. 35. 



Now 


since 






j! 




RS 11 PQ, 




[4] 


and 


RH II PK, 


by [i] 


[5] 


the pla 


ne RSLH is || to the plane PQMK. 








by Geom. 41. 


[6] 


Hence 


SL II QM. 


by Geom. 39. 


[7] 


Draw 


HU_LRS and KT _L PQ. 




[8] 


Now 


RS XSIv " PQ _L QM. 


by Geom. 33. 


[9] 


Hence 


HU II SL " KT II MQ. 


by Geom. 46. 


[10] 


Hence 


by [6] HU II KT. 


by Geom. 37. 


[II] 


But 


RS II PQ II KM II HL. 


by Geom. 35. 


]l2" 


Hence 


HU = US and KT --= MQ. 


by Geom. 17. 


Now since 






[13] 




RH II PK, 


by [i]. 


>4; 


and 


HU II KT. 


by [10]. 


\\^'_ 


Then 


z^RHU = ^PKT. 


by Geom. 41. 


;i6; 


Now 


HU = RH cos RHU, 


by Trig. 2. 


[17] 


and 


KT = PK cos PKT. 


by Trig. 2. 


[is; 


But 


RH = PK, 


by Geom. 17. 


[19] 


and 


cos RHU = cos PKT. 


by [15; . 



Hence by [16], [17], [18] and [19] 
[20] HU = KT. 

Hence by [12] 
[21] US = QM. 

Now since by [6] and [21], LS and MQ are both equal and 
parallel, SQMUis a parallelogram. by Geom. 48. 

Hence BQ || DM. 

Q. K. D. 



CHAPTER III 

Transformation of Coordinates 

PROPOSITION I 

406. If we have given the coordinates of a point referred to any 
system of rectangular axes we can find the coordinates of the 
same point referred to any other system of axes parallel to the 
first by putting 

X -:=■ m -\- x\ 
y z= n -\- y\ 
2- = -{- z\ 

in which x, y, z are the coordinates of the point referred to the 
original axes, afid m, n, the coordinates of the new origin 
referred to the original axes, and x\ y\ z' the coordinates of the 
point referred to the new axes. 



312 



ANALYTIC GEOMETRY 



In Fig. 149 let OX, OY and OZ te the old axes, and 
0'X^ O'Y' and O'Z' be the new axes. 

Let P be any point in space and draw its coordinates with 
respect to each system of axes. 

Let X ~ OH, J = HK and ^ = PK, 

x' = O'R, y = RQ " 2' = PQ, 




Fig. 149 

Now using the notation of § 400, we have 



"l[ 


OPx = OO'x + O'Px . by § 403 


[2] 


But OO'x — OLx + LMx + MO'x , by § 403 


'X 


and 0'Px=0'Rx+RQx+PQx. by § 403 




Hence 


[4] 


OPx = OLx + IvMx + MO'x + O'Rx + RQx + PQx 


[5] 


But OPx=OHx + HKx+PKx. by §403 


[6] 


Hence OHx + HKx + PKx = 



OLx + LMx + MO'x + O'Rx + RQx + PQ. 



TRANSFORMA TION OF COORD IN A TES 313 

[7] Hence OH cos 0° 4- HK cos 90'' + PK cos 90° =^ 
OLcos o'' + IvM cos 90° + MO' cos 90° + O'R cos 0° + 



RQ cos 90"" + PQ cos 90' 

[8] Hence OH = OIv + O'R, 

[9] or X =^ m-\- x'. 

[10] Similarly y ^=: n +jf', 

[11] and 2 =1 -{- s'. 



by §404. 
by Trig. 19, 



Q. K. D 



PROPOSITION II 

407. If we have given the coordinates of a point referred to 
any rectangular system of axes, we caji find the coordinates of the 
same point referred to any other rectangular system having the 
sa7ne origin by putting 

X z=. x' cos a -\-y' cos a' -{- 2' cos a", 
y ^:i x' cos fi -{- y' cos ^' -\- 2' cos /3", 
2 z= x' cos y -\- y cos y + 2"' cos y", 

in which x, y, 2 are the coordinates of the point referred to 
the original axes ; x' y', 2' its coordiyiates referred to the new 
axes; the a' s are the angles which the neiv axes make with the 
original X axis ; the fi' s the angles which the new axes make 
with the original Y axis ; and the y^s the angles which the 
7iew axes make with the original Zaxis. 




Fig. 150 



\u 



ANALYTIC GEOMETRY 



In Fig. 150 let OX, OY and OZ be the original axes, and 
OX', OY' and OZ' the new axes. 

Let P be any point in space and draw its coordinates with 
respect to each system of axes. 

Let;^=OH, y=HK and -^ = PK, 
jr'=OM, y=MIv " y = pi„ 

and in the following table let the angle between any two axes 
be in both the column and the horizontal line containing 
those axes. 





OX 


OY 


OZ 


OX' 


a 


/i 


y 


OY' 


a' 


f^' 


y' 


OZ' 


a" 


fr 


y" 



Thus a is the angle between OX' and OX, 
and y" " " " " OZ' " OZ, etc. 

Now in Fig. 150 



[i] OHx+HKx+PKx=OMx+MLx+LPx. 

by § 403. 
[2] Hence OH coso°+ HK cos 90° -|- PK cos 90" =1 

OM cos a + ML cos «:' -|- LP cos a" . by § 404. 

[3] Hence x = jf ' aos (x -\- y qo^ a' -\- z\q.os a" . by Trig. 19. 
Similarly by projecting upon the Y axis, we get 

[4] jj/ = ycos/?+jj/'cos /5' + y cos/^". 

And by projecting upon the Z axis, we get 

[5] z^=^ x' cos y -\- y cos y + z' cos 7". 

o. E. D, 

408. Corollary. — If zve project the coordinates of the point upon 
the' new axes, we gel 



[6] 
[7] 
[8] 



X 



X cos oc -\- y COS i5 -|- z cos Y, 

y' z=i X COS a' + y COS ;3' -\- Z COS y\ 

2'' ;= X COS <a^"+_j^ cos/5"-|-^ cos;-". 



TRANSFORMA TIOX OF COORDIXA TES 



315 



Equations [3], [4], [5], [6], [7] and [8] may be con- 
venientl}^ indicated b}' means of the following 



TABLE. 





x' 


y' 


z' 


X 


COS oc 


COS a' 


COS «'" 


y 


COS iJ 


COS /5' 


cos /5" 


z 


COS Y 


cos r' 


COS /'" 



Equations [3], [4], [5], [6], [7] and [8] ma}' be readih- 
obtained from the table above b}" means of the following rules : 

409. Rule I . — The variable at the left on any horizontal line 
is equal to the sum of the products obtained by ?nultiplying each 
cosiiie on that line by the variable at the top of the column con- 
taining the cosine. 

410. Rule 2. — The variable at the top of a7iy cohon^i is equal 
to the sum of the products obtained by multiplyi^ig each cosine 
in that column by the variable at the left on the horizontal lijie 
containi7ig the cosine. 

411. Corollary i. — cos^ a -f- cos'^ /5 -j- cos' y =1 i^ 

cos'^ a' -f- cos^ /5' + COS' r'= ^, 

cos' a" + cos' ;S" + cos' y" = z. by § 383. 

412. Corollary 2. — cos a cos a' -f- cos /5 cos ^3' -|- cos y cos y' zzz o. 

cos a cos a" -f- cos 3 cos ;5'' + cos y cos y" := o . 
cos oc' cos a" -f- cos, 'j' cos h" -\-cos y' cos y" =0. 

by § 395. 



CHAPTER IV 

Spherical Coordinates of o Point in Space 
Z 




Fig- 151 

In Fig. 151 let P be any point in space, and let 

(p = QOX, = POQ and p = OP. 

Then if we know cp, and p, we know the position of the 
point P. 

413. The Radius Vector. — The radius vector oi a point in 
space is its distance from the origin. 

414. The Latitude of a Point. — The latitude of a point in 
space is the angle between its radius vector and the projection 
of the radius vector upon the XY plane. 

415. The Longitude of a Point. — The lojigitude of a point 
is the angle betw^een the X axis and the projection of the 
radius vector upon the XY plane. 

416. The Spherical Coordinates of a Point. — Theradius 
vector, the latitude and the longitude of a point are called its 
spherical coordinates. 



SPHERICAL COORDINATES 317 

PROPOSITION I 

417. If we have given the rectangular coordinates of a pointy 
we can find its spherical coordinates by putti?ig 

jT =: p COS cp cos 6, 

y ^=. p COS cp sin 0, 
2 -=z p sin &. 
lyet X = OH, y = HK and 2 = PK. 

Then in Fig. 151, since the axes are rectangular, KHO is 
a right angle. 

OH = OK cos KOH, by Trig. 2. 

X = OK cos (p. 
0K = p cos 0. by Trig. 2. 

X ^=1 p cos (p cos 0. 
KH = OK sin KOH, by Trig. i. 

y = OK sin q). 
OK = p cos 6. by Trig. 2. 

y ^=: p sin (p COS 0. 
PK = OP sin d, by Trig. i. 

2- = p sin 6. 

418. Corollary. — 1/ cos a^ cos fd, cos y are the direction cosines 
of the radius vector of a point in space, theji 

cos a ^ cos cp cos 6, 

cos (3 = cos cp sin 6, 

cos y = sin 8. 



'_1_ 


Hence 


[2] 


or 


[3] 


But 


[4] 


Hence 


[5] 


Again 


>; 


or 


[7] 


But 


[8] 


Hence 


'.9. 


Again 


[lol 


or 



For 


in Fig. 151 






[i] 




0H=: 


OP cos POH, by Trig. 2. 


[2] 


or 


X = 


p cos a, 


[3j 


and 


X = 


p cos (p cos 6. by §417, [4]. 


[4] 


Hence 


cos oc = 


cos cp cos 0. 


[5] 


Similarly 


y = 


P cos POH z= p cos /3, 


[6] 


and 


y — 


p sin (p cos 6. by §417, [8]. 


[7] 


Hence 


COS ^ = 


sin cp cos 0. 


[8] 


Also 


cos y = 


sin d. 



: CHAPTER V 

The Plane 

PROPOSITION I 

419. The equatio7i of a plane is 

X cos a-^ y cos fi -\- 2 cos y — /> = o, 

171 which x,j/, z are the coordinates of any point in the plane ^ 
p the length of the perpendicular drawn from the origin to the 
plane ^ and cos a, cos /?, cos y the direction cosines of this perpen- 
dicular . 




Fig. 152 

lyCt MN be any plane, and OS a line drawn from the origin 
I to this plane at the point R. 

Let P be any point in space and draw its coordinates PK, 
KH and OH. ' 

Draw PQ _L to the plane MN. 
Let a^ /5, y^ be the angles which OS makes with the axes. 

Let/ = the length of OR. 

Let PKs = the length of the projection of PK upon OS. 



THE PLANE 319 

Praject the two broken lines ORQP and OHKP upon the 
line OS. 



[i] Then OHs+ HKs + KPs= ORs+ RQs+ QPs. 

by § 403. 
[2] Hence x cos oc -\- y cos P -\- 2 cos y = 

p cos 0° + RQ cos 90° + ^cos o^, by § 404. 

[3] or .r cos <a'-|-jj/ cos/? + ^-cos ;k ^=/ + (^. by Trig. 19. 
[4] Hence .rcos <a' +JF cos /? -j- -s* cos }^ — pz=^d. 

If the point P is taken anywhere in the plane MN, then 
^z= o, and [4] becomes 

[5] X cos oc -|- jv cos (^ -\- 2 cos y — p = o. 

Now cos oL^ cos /5, cos y , and p determine the point R. 

And there can be but one plane drawn through R perpen- 
dicular to/. by Geom. 53. 

Hence since there can be but one plane whose perpendicular 
from the origin has the length p and whose direction cosines 
are cos 01., cos />, cos «', [5] determines the plane MN. 

Since the x, y and z of [5] are the coordinates of any point 
on this plane, then [5] must be the equation of that plane. 

Q. K. D. 

420 . Co roll a ry . — 

The equation of a plane \ to the XY plane is z ■=^ p. 

ZX " '' y=p. 

' ZY " " X =p. 



( 1 ( ( ( ( II ( ( ' < 



( < k < i ( 



PROPOSITION II 

421. Every equation of the first degree contaifiing three varia- 
bles only is the equation of a plane. 

[i] I.et Ax + By + Cz-\- D = o, 

be an}' equation of the first degree containing only three vari- 
ables. Then if this be the equation of a plane, it must be 



320 ANALYTIC GEOMETRY 

but another form of the equation 

[2] Ji: cos « -(- J/ cos /^ + 2- cos / — p = o. by § 419. 

Therefore by muUiplying [i] through by some constant 
as /V, we must be able to change [i] into the form of [2]. 

[3] That is XAx + XBy-{-XC2 + XD=^o, 
must be only another form of [2] . 

[4] Hence XA = cos a, 

XB = cos /?, 
XC = cos y, 
XD=—p. 

Now since A^, XB and XC are the direction cosines of a 
line through the origin, then 

[5] AM^ + A^^^ + rC^^i, ■ by §383. 

[6] or X\A' + B''+C')=i. 

[7] Hence A. = ± — ■. 

V A' + B' + C 

Since A, B and C are real quantities, A. must be real. 
Substituting the positive value of A. into [4], we get 

■ [8] 

A B . 

= cos <T ; — ^^:i:zz=^z=^^^ = cos p ; 



V A' + B' + C V A' + B'' + C 

cos y ; — zzi^^^ziiz^i^zr = — P- 



V A' + B' + C V A' + ^' + C 

Substituting the positive value of A. into [3], we get 

[9] — z=-^+ ^ — -y 

V A'' + B' + C V A' -\- B' + C 

. C , D 

• V A' -\- B' ^ C V A' + B' + C 

Hence from [8] and [9], we get 



THE PLANE 321 

[10] JT cos <^ + jK cos ^ + -^ COS ;k — / = o. 

Therefore [i] is only another form of [2], and hence [i] is 
the equation of a plane. 

Q. E.D. 

From [8] we get the following 

422. Corollary. — If Ax -f- By -\- Cz -\- D z=iO be the equation 
of a plane, p the J_ on it from the origin, and a, y5, y the direc- 
tion angles of this J_, then 

/ = — , cosa=. 



V A' + B' -^ C V A' + B' + C' 

COS p = — - and cosy=i- 



V A' ^- B' + a V A' -^ B"- -{- c 



PROPOSITION III 

423 . If X cos a -\- y cos f^ -\- z cos y — /> = o 
be the equatioji of a plane, then 

X cos a -\- y cos fJ -{- z cos y — ^ ± ^ 1= o 

is the equation of a plane pa^'allel to it. 

For since the direction cosines of the J_ drawn from the 
origin to the two planes are the same, by § 419. 

these J_'s must coincide with each other. Hence the two 
planes are JL to the same straight line and are therefore || . 

by Geom. 38. 

Q. K. D. 

424. Corollary. — The distance between the two planes is dz d. 

For the distance from the origin to the first plane is p, and 
the distance from the origin to the second plane is/±^. 
Hence the distance between the planes must be ± ^. by § 419. 



322 ANALYTIC GEOMETRY 

PROPOSITION IV 

425. The distance frof-n any point x\ y' , z' to the plane 

X cos a -\- y cos ft -\- z cos y — p :^ o 

is 

x' cos a -{- y COS f3 -\- 2' cos y — p. 

For let X cos oc-\- y cos ft -\- z cos y — p =: o . 
be the equation of any plane. 

Let x\ y z' be the coordinates of any point in space. 
Let d = the distance from x' , y\ z' to the plane. 

Pass a plane through the point x' , y\ z' || to 

x cos a -{- y cos ft -\- z cos y — p = o. 

Since these || planes are everywhere equall}^ distant, ^ is the 
distance between them, and the equation of the second plane is 

[ I ] X cos f^ -\- y cos ft -\- z cos y — p^ d^ o. 

Now since the point x', y', z' is on the second plane, its co- 
ordinates must satisfy the equation of that plane. 

Hence substituting x',y\ z' into [i], we get 

[2] x' cos oc-\- y' cos ft -\- 2' cos y — p d= ^ = o. 

[3] Hence ± d ^=^ x^ qos> oc -\- y' cos ft -\- z' cos y — p. 

Q. E. D. 

By §§ 422 and 425 we can easily prove the following 

426. Corollary. — If d = the dista^ice from the point x\ y\ z' to 
the plane 

Ax-\-By + Cz + D = o, 

^ Ax' + By+Cz' + D 
then d = ' -^ ' ' . 

1/ A' + B'+ C 

427. The Traces of a Plane. — The traces 0/ a plane are its 
intersections with the three coordinate planes. 




323 



Fig- 153 



In Fig. 153 AB is the trace of the plane MN on the plane 
ZX, 

BC is its trace on the plane XY, 

and AC is its trace on the plane ZY. 

Let the equation of the plane MN be 

[i] Ax^By+Cz -^ D^o. 

Now the plane MN cuts the plane XY where ^ = o. 
Hence making 2- = o in [i], we get 
[2] Ax^ By^ D=o. 

for the equation of BC the trace of MN on XY. 
Similarly making j/ = o, we get 
[3] Ax^Cz -^D^o, 

for the equation of the trace on ZX. 

[4] And making ;tr =1 o, we get 
[5] By-YCz^D = o, 

for the equation of the trace on ZY. 



324 ANALYTIC GEOMETRY 

428. Scholium. — The variables in the equation of a trace 
indicate the coordinate plane on which the trace lies. 

429. Intercepts of a Plane. — The distances along the axes 
from the origin to any plane are called the intercepts of that 
plane. 

In Fig. 153 the intercepts are OA, OB, OC. 

The coordinates of the point B are :ir = OB, y^^o, and 

Since B is on the plane MN, its coordinates must satisfy 
the equation of that plane. 

Hence substituting jr = OB, j/ = o, and ^ = into the 
equation 

Ax-\- By^Cz-^ D — o, 
we get 

[i] ^(OB) + i9 = o. 

[2] Hence OB = -j = the intercept on the X axis. 

A 

[3] Similarly OC = ^ = the intercept on the Y axis. 

Jd 

[4] and OA = -^ = the intercept on the Z axis. 



CHAPTER VI 

Straigfht Lines 



PROPOSITION I 



430. The equations of a straight line in space are 
X — x' y — y' z — 2' 



cos OL 



COS f^ 



COS y 



in which x' , y' z' are the coordinates of a fixed point on the line ; 
x^ y, z are the coordinates of any other pohit on it ; a?id cos a, cos /?, 
cos y are the direction cosines of the line. 




Fi.?-. 154 

IvCt AB be any line in space. 

lyCt A be a fixed point on this line, and P be a7iy point on it. 

Let the coordinates of A be x', y' z\ and 

" '' " " P " x,y, z. 

Through each of the points A and P pass a plane _L to OX. 
Then the plane RS is || to the plane PQ. by Geom. 38. 



326 



ANALYTIC GEOMETRY 



Draw AK || to OX. 

Then AK is _L to the plane PQ, 
and AK_LtoPK, 

and the Z- AKP is a right angle. 



Through the origin draw 



by Geom. 36. 
b}' Geom. 33. 



OH II AB. 
[1] Then z^LOH = z^KAP. by Geom. 11. 

I^et M be any point on OH and through it pass planes _L 
to the axes. 

The coordinates of M will be OL, LN and MN. 
Join M with U, L and W. 

OIv is _L LM. by Geom. 33. 

Hence Z. OLM = right Z.. 

Therefore by [r] the two triangles OLM and AKP are 

by Geom. 51. 

AK_ AP 
Ol7"" OM" 

AK= SQ. 

SQ _ AP 



similar. 

[2] Hence 

[3] But 

[4] Hence 



by Geom. 31. 
by Geom. 40. 



[5] or 



OIv OM' 

x — x' AP 



OL 



OM* 



It may easily be shown that PJK and MNI^ are similar. 

by Geom. 51. 

JK PK AP 
[6] Hence ^n^^ML-QM' 

y—y'_ AP 



by Geom. 31. 



[7] or 

[8] Similarly ^^ 



LN OM 

z — z' _ AP 
' OM" 



STRAIGHT LINES 327 



x — x' y — y' z 
[9J Hence — 



OIv I.N MN 

Now let OM be taken as the unit of length. Then OL, 
LN and MN are the direction cosines of the line, and [9] be- 
comes 

^ _, X — x^ y — y z — z' 

[10] -■" -^ - 



cos a cos (i cos y 

Since x, y, z are the coordinates of an}^ point on the line 
AB, and the other quantities in the equation are constants, 
[10] must be the equation of the line. 

Q. E. D. 

PROPOSITION II 

431. The equations of a straight line may be written 

X ^=^ x' -\- Ip, 
y =1 y -\- ?np, 
z := z' -\- np, 

in which x\ y' z' are the coordinates of a fixed point on the line ; 
X, y, z the coordinates of any point on the line ; p the distance 
between these points ; ajid I, m, n the directors of the li7ie. 

In Fig. 154 

[i] OL = OM cos a^ 

[2] LN = OW = OM cos /?, 

[3] MN =1 OU = OM cos y. by Trig. 2. 

Let/ be any constant whatever, then 

[4] j^^.OL =/.0M cos o', 

[5] /.LN =/.0M cos A 

[6] /).MN = /.OM cos y. 

Hence from [4], [5] and [6], we get 

|- -. /.OL _ ALN _ /.MN 

cos a ^~ cos f^ cos y ' 



3^8 



ANALYTIC GEOMETRY 



[8] Let 

[9] 
[10] 

[II] 



d^p.OU. 



Then by [7], [8], [9] and [10], we get 

P -, a b c 

[12] 



cos a cos/^ cos y' 

Let p = AP. 

Then from § 430 [9] and the similar triangles KAP and 
LOM, we get 

J- -. X — x' y — y' z — z' p 

'-^^-' 01. "~ LM ~ MN "" OM* 

Dividing this equation through by/, we get 



X — X 



y —y 



P 



^^^^ p.Oh p.LM />.MN /.OM ■ 

Hence by [8], [9], [10] and [11], [14] becomes 



ri.i -^ -^ 


y — y z — z 


- _ 


I 


L'^J a 


b c 


"16] Hence 


. . a 
x = x' + ^ p. 






-^7; 


y^y ^ d ^' 






[18] 


z =^ z' + -^ p. 






Now let / E 


a b 

- d' "" - d' " 

a 


^ 


C 
- d 


;i9] Then 


Ida cos 


a 




m b b cos 


p' 






d 
b 






"20] and 


7)1 d b cos 

71 C C COS 

d 


I 

r' 





by [12]. 



by [12]. 



STRAIGHT LINES 329 

Then from [19] and [20], we get 

^ -, I VI n 

[21] = J.— . 

cos a cos p cos y 

Hence /, m and n are the directors of the line AB. by § 391 . 

Substituting /, m and n into [16], [17] and [18], we get 

[22] X ^^ x' -\- Ip, 

[23] _ ' y — y -\- mp, 

[24] 2" = y -f" 7lp. 

Now in Fig. 154, p is the distance along the line AB from 
the given point x\ y\ 2' to the point P. If then the point P 
be known, the value of p is known. Substituting this value 
of p into [22] , [23] and [24] will give us one and only one 
set of values for x, y and z. 

On the other hand, if the values of x, y, 2 are given, then 
[22], [23] and [24] will give us the length of p. Measuring 
off this length from the point x' , y' , 2' in the direction of the 
line AB, we get one and only one point on the line. 

Hence for each point P on the line AB there is one and only 
one set of values of x, y, and 2 that will satisfy [22], [23] and 
[24] ; and for each set of values of x,y and 2 satisfying these 
equations there is one and only one point on the line AB. 

Therefore 

X = ^' + /p, 

2=^ 2' -\- np, 

determine the line AB and are called the equations of the line 
AB. 

PROPOSITION III 
432, The equations of a straight line may be written 

[i] x^=^sz-\-b, 

[2] and y^=isz^-^h'. 

Equation (z) being the equation of the projection of the liiie 



330 



ANALYTIC GEOMETRY 



upon the ZX plane, s its slope upon the Z axis, and b its inter- 
cept on the X axis. 

Equation {2) being the equation of the projection of the line 
on the ZY plane, s' its slope on the Z axis, and b' its intercept on 
the Y axis. 




Fig- 155 

Ivet AB be any line in space. 
" DC be its projection on the plane ZX. 

Let s = tan ZEC, 
b= DO, 
5' = tan HJO, 
/^' = H0. 

Through the origin draw OM || to AB and a unit in length, 
and let its direction cosines be cos «', cos ^, cos y. 

Let OQ be the projection of OM on the ZX plane. 
Through M pass planes J_ to the axes, 
[i] Then OL = cos <:f, LN=3COs/5, MN = cos;k. 



STRAIGHT LINES 331 

r -1 XT OL IN SQ ^ ^ 

W Now MN^lfN^OS- byGeom.47. 

[3] But -^^=:tanQOS. 

[4] Hence ^ = tan QOS. 

Now OQisllDC. by §405. 

[5] Hence tanQOS = tanZEC=: ^. byGeom. 11. 

Hence by [4] and [5] , we get 

^^^ MN - ^• 

[7] Similarly -^^ — s'. 

Now by § 430, the equation of the line AB is 

[8] ^—^ = y^ = ^^i'. 

cos (^ cos p COS y 

Hence by [i] and [8] we get 

X — x' y — y z — 2' 

•-^^ OIv ~ IvN ~ MN • 

Now let x', y, z' be the point A where AB pierces the XY 
plane. 

[10] Then x' = AH = DO = b, 

[11] y = AD = HO = ^', 

[12] and z' ■=. o. 

Substituting these values of x\ y , z' into [9], we get 
X — b y — V z 



[13] 



OIv LN MN 



[14] hence "^'^MN'^"^^" 

Hence by [6] we get 

[15] x — sz^b. 



132 



ANALYTIC GEOMETRY 



Now if we consider OZ as the axis of abscissas, and OX 
as the axis of ordinates, [15] will be the equation of the 
line DC. b}^ § 53. 

From [13] we also get 

LN 



[16] 



y 



MN 



z-^b\ 



Hence by [7] we get 

[17] y=s'z^b\ 

which is the equation of KH. by § 53. 

Now one plane and only one can be drawn _L to the plane 
ZX, which shall contain the projection DC. by Geom. 45, 

This plane will contain the lines DA andCB. by Geom. 42. 
Hence since it contains the points A and B, it must contain 
the line AB. 

Similarly it may be shown that the only plane that can be 
passed through the projection KH _L the plane ZY must also 
contain the line AB. Hence AB must be the intersection of 
these two planes. 




Fig. 156 



STRAIGHT LINES 333 

Now since in Fig. 156 the only plane that can be passed 
through the projection DC perpendicular to the plane ZX 
must contain the line AB, and the only plane that can be 
passed through the projection KH perpendicular to the plane 
ZY must also contain the line AB, then the projections DC 
and KH determine the line AB. Therefore the equations of 
the projections DC and KH, namely, 

y—s'zArb\ 

determine the line AB, and are called the equations of AB. 

Q. E. D. 



CHAPTER VII 

Surfaces 
PROPOSITION I 
433. A surface may be represented by an equation of the form 

f{x,y, z) = o. 




Fig- 157 



In Fig. 157 let MN represent any surface, and let OX, OY 
and OZ be the axes of coordinates. 

From any point P on the surface draw PK JL to the XY 
plane. It will generally cut the surface in other determinate 
points, as P', P", etc. 

Draw KH || OY, 

and' KJ II OX. 

Let X = OH, 
J/ = KH, 
^ = PK, 
2' = P'K, 
2'" = P"K. 



SURFACES 335 

Now to the value of 2 belonging to each point on the sur- 
face there corresponds one and only one value of x, and one 
and onh' one value of y. That is, the coordinates of every 
point on the surface have a definite relation to each other. 
This relation may be expressed b}" the equation 

/(^,jj/, 2) =0, 

which is therefore called the equation of the surface. 

434. The Equation of a Surface. — The equation of a sur- 
face is one in which the variables represent the coordinates of 

every point on the surface. 

PROPOSITION II 

435. To find where a straight line cuts a siirface^ we must treat 
the equations of the straight liiie a7id of the surface as simulta- 
neous and solve them. The values of x, y and z thus found are 
the coordinates of the cutting points. 

In Fig. 157 let MN represent any surface and RS any 
straight line cutting it at the points R, R', R". 

[i] IvCt /(jr, J, 5") =0, 

be the equation of MN, 

[2] and let ^ = .r' + /p, 

[3] y —y' + ^p. 

[4] z — 2' -\- np, 

be the equations of the line RS. by § 431. 

Let The any point on the line RS, whose coordinates x\ 
y\ 2"' we know. 

Then by § 431, x\ y', z\ /, m, n in [2], [3] and [4] are 
known and p = TR. 

Now since each of the cutting points, as R, is on the sur- 
face, the coordinates of R must satisfy [i] . 

Since the point R is also on the line RS, its coordinates 
must also satisfy [2], [3] and [4]. 



336 ANALYTIC GEOMETRY 

Therefore let x, y and z represent the coordinates of the 
point R. Then [i] , [2] , [3] and [4] are simultaneous and 
may be solved. 

Hence substituting the values of x, y, z given in [2], [3] 
and [4J into [i], we get an equation 

/(x' + Ip, y + mp, z' + 7zp) = o, 

whose only unknown quantity is p. 

Solving the equation 

[5] /{^' + ^p, y + ^^2p, ^' + np) — o, 

we get the values of p, that is TR, TR', TR", etc, The num- 
ber of values of p depends upon^'the degree of [5]. 

Substituting each of these values of p successively into [2] , 

[3] and [4], we get the coordinates of the points R, R', R", 

etc. 

O. K. D. 

436. Corollary. — To find where two surfaces cut each other we 
must treat the equations of the surfaces as shnultaneous and solve 
them. The values of the variables thus fozmd a7^e the coordinates 
of the cutting points. 



CHAPTER VIII 

Quadfics 

437. The Quadric. — The locus of every equation of the 
second degree containing three variables only is called a 
quadric. 

PROPOSITION I 

438. A straight line intersects a quadric i7i two real^ imagi?i- 
ary or coincident points . 

The general equation of the second degree containing three 
variables onl}^ is ^ 

[i] Ax' +Ay"' + A" 2' + 2By2+2B'zx+ 2B" xy + 2Cx 

By § 437 every quadric can be represented by some form of 
this equation. 

The equations of the straight line are 
[2] jc = .r' + /p, 

[3] y ~y' + ^P) 

[4] z— z' -^ np. by §431. 

Now to find where the straight line cuts the quadric, w^e 
treat [i], [2], [3] and [4] as simultaneous and solve them. 

by §435. 
Hence substituting the values of ^, jK and z given in [2], 
[3] and [4] into [i], we get an equation of the form 

[5] A{x' + Ip) + A'iy + ^^P) + ^"(^' + np) 
+ 2B{y + mp) {z' + np) + etc. 

Since in [2], [3] and [4] x\ y' z\ /, m and n are known 
quantities, p will be the only unknown quantity in [5]. 
Again [5] contains the second and no higher powers of p. 
Hence when we solve it we get two and only two values for p. 



338 



ANALYTIC GEOMETRY 



Substituting these two values of p successively into [2], 
[3] and [4] , we will get two and only two sets of values of 
;r, J/, z. But each of these sets of values of x, y, z locates a 
cutting point. 

Hence the straight line cuts the quadric in two points and 
two only, which will be real or imaginary according as the 
values of x, y and 2 are real or imaginary, and will be coin- 
cident when the two sets of values are identical. 

Q. K. D. 

439. A Chord. — If a straight line cuts a quadric in two 
points, the part of the line joining these points is called a 
choi^d of the quadric. 

PROPOSITION II 

440. Every section of a quadric made by a plane is a conic. 




V 'r 



Fig. 158 



QUADRICS 339 

For in Fig. 158 let the curved surface represent a quadric, 
and let 

[i] Ax" + A'y" + A"z' + 2Byz + 2B'zxr^ iB^'xy + 2Cx 
+ 2C'y+2C"z + D — o, 

be the equation of that quadric referred to the axes OX, OY 
and OZ. Then to get the equation of the quadric referred to 
any other system of axes OX', OY' and OZ' having the same 
origin, we must put 

[2] X = x' cos (X -{- y' cos a' -\- z' cos oc" , 

[3] y z= x' cos /? -\- y' cos /5' + 2' cos ji" , 

[4] z ^=. x' cos y -\- y' cos ;k' + z' cos /", 

into [i]. by § 407. 

[5] Let oi =: cos a, 

[6] /3 = cos /i, 

[7] r = cos r, 

ex' ^ cos ^', 
etc.=^etc. 

in [2], [3] and [4] and substitute the results thus obtained 
into [i] . We will then get 

[8] A{ax' + a'y' +a"z'r -\- A' ifdx' + /3'y' -^ /3"z'y 

+ A"i r 

+ 2B{/3x' 4- /3'y' + ^"z') iyx' + v'y' + y"z') 

+ 2B'{ )( ) 

+ 2B"( )( ) 

+ 2C(ax' + ay + a"z') + 2C'( ) 

+ 2C"( ) +I) = o. 

[9] Now {ax'+a'y'-\-a"z'y = 
a^x''' -|- ay + a^"z'^ + 2«'«'^>' + 2aa"x'z' + 2a'ayz\ 

[10] and (yS;r' + ySy+^V)(r-r' + ry + r"^') = 

It will be found that when the terms in [8] are expanded as 
in [9] and [10] and the result is factored with respect to Jr'^ 
y''\ y^ x'y' , x'z\ x',y\ and^, we get an equation having ex- 



340 ANAL YTIC GEO ME TR Y 

actly the same form as [i], which may therefore be written 

[i I J ax''' + a'y'^ + a" ^4- 2by' z' -\- 2b' z' x' + 2b"x'y 
+ 2cx' + 2c'y' -\- 2c"z' + Z^ = o. 

The values of the coefficients a, b, c, etc., in this equation 
will depend upon the values of «', ft, y, and A, B, C, etc. in [8] . 

Now in Fig. 158 let M'N' be a plane 1| to the plane X'Y', 
that is, the plane MN. 

The equation of M'N' will be 

[12] z =^ p. by § 420. 

Since OX', OY' and OZ' may have any directions, the 
plane MN may slope in any direction. 

Hence since M'N' is parallel to MN, and p may have any 
value, M'N' may represent any plane. 

To find where the plane M'N' cuts the quadric, we must 
treat [11] and [12] as simultaneous and solve them. § 436. 

Hence substituting the value of z found in [12] into [11], 
we get 

[13] ax'' + a'y"+ «"/ + 2bpy + 2b'px' + 2b"xy' 

-f- 2CX' -\- 2c'y -(- 2c"p -{- D ^=0. 

Now since this is an equation of the second degree contain- 
ing two variables only, it is the equation of a conic, by § 33i<2. 

Hence in Fig. 158 RS, the common section of the plane and 

quadric, is a conic. 

Q. K. D. 

441. The Center, — The center of a quadric is the point 
which bisects every chord passing through it. 

442. A Diameter. — A diameter of a quadric is any chord 
passing through the center. 



QUADRICS 341 

PROPOSITION III 

443. The eqjiaiiou of a quadric referred to its center is 
Ax' + A'y' + A" 2' + 2By2 + 2B' zx + 2B''xy\-D' — o. 

For [i], § 440, is the equation of the quadric referred to the 
axes OX, OY and OZ of Fig. 158. 

Now to get the equation of the quadric referred to the || sys- 
tem 0"X", 0"Y" and 0"Z", we must put 

[14] X =^ m -\- x\ 

[15] J/ =^ n +y, 

[16] 2=10 -\- 2', by §406. 

into [i]. Making this substitution, we get 

[17] Ax" + Ay + A" 2" + 2By2' + 2B'2'x' + 2B"xy 
+ 2{Am + B''n + B'o+ C)x' 
+ 2{Ahi-\-Bo-^B''vi+ C')y' 
-\- 2{A"o + B'7n+ Bn + C")2' 
-\- Am^ -\- A'n^ + A"o^ + 2Bno + 2B'om + 2B"mn + 2C7;z 

Now suppose the new origin O" in Fig. 158 to be such that 
when the quadric is referred to the axes 0"X", 0"Y", 0"Z", 
the coefficients of x\ y' and 2' in its equation become o. 

Then from [17] we get 

[18] Am + B"n + B'o + C =: o, 

[19] A'u +Bo + B"7n + C = o, 

[20] A"o + B'm+ Bn ^ C" — o, 

or 

[21] Am + B"7i -\- B'o = — C, 

[22] B"m+A'7i -\- Bo = — C\ 

[23] B'm + Bn + A" — — C" . 



342 



ANALYTIC GEOMETRY 
—C B" B 
—C A' B 
—C B A" 



n 



_1-IV,V_ 


ffi- — 


A 


B" 


B' 






B" 


A' 


B 






B' 


B 


A'' 


A 


— C 


B' 






B" 


c 


B 






B' 


C" 


A' 






A 


i5" 


B'' 







B" 


A' 


B 






B' 


B 


A" 







A 


B" 


—C 


B" 


A' 


—C 


B' 


B 


C" 


A 


B^' 


B' 


^" 


A' 


B 


B' 


B 


A" 



Substituting these values of m, n and o into [17], we get 
by [18], [19] and [20], 

[25] Ax" + A'y" + ^'V^ + 2By'z' + 2B' z' x' 
+ 2B"x'y' -\-D' = 0, 
in which 

[26] D' = Am- + A'n' + A"o' + 2Bno + 2B'mo + 2B"mn 
' + 2Cm + 2C'n + 2C"o + I?. 

Since accents over the variables are no longer needed, we 
may drop them from [25], when we will get 

[27] Ax' + Ay + A''2' + 2By2+2B'x2+2B"xy + T?'=o, 

which is the equation of the quadric referred to the origin O". 

Now the form of [27] shows that if x, y, z satisfy it, then 
— -^j — J/ and — z must also satisfy it. That is, if any point 
x,y, 2 be on the quadric, then the point — x, — y, — z will 
also be on it. But these points are on opposite sides of the 
origin and equally distant from it. 



QUADRICS 343 

Hence the origin O" bisects every chord passing- through it 
and must therefore be the center of the quadric. b}' § 441. 

Hence [27] is the equation of the quadric referred to its 
center. 

Q. E. D. 

In [24] m^ n and are the coordinates of the origin O". 

by § 406. 

But we have just shown that O" is the center of the quadric. 
Hence m, n and are the coordinates of the center. 

444. Scholium. — The right hand member of [26] is a func- 
tion of 7n^ n and of exactly the same form as the left hand 
member of [i] , § 440. 

By [24] we see that the coordinates of the center will be 
real. That is, the quadric will have a center when the de- 
terminant which forms the denominators of the fractions is not 
o, and that these coordinates will be infinite, that is, the 
quadric will have no center when this determinant is o. 

445. Central Quadrics. — Quadrics which have a center 
are called central quadrics. 

446. Non-Central Quadrics. — Quadrics which have no 
center are called non-central quadrics. 



CHAPTER IX 

Central Quadrics 

PROPOSITION I 
447. The equation of the central quadric may be written 

rt -b^^d- = I 

— I — 9 — I — 7 9 — I 9 -■• • 

a b^ c 

For [27] is the equation of the quadric referred to the axes 

0''X", 0"Y" and 0"Z" in Fig. 158. 

To get the equation of the quadric referred to the axes 
0"X"', 0"Y'", 0"Z"' we must, as in [8], put 

X = ax' + a'y' -\- a" z' , 

y = fSx' + /?y + P"2\ 

2 = yx' + yy' + /"^', 

into [27]. 

Now it may be shown, as it was for [8], that this will give 
us another equation of exactly the same form as [27] . 

Hence this new equation may be written 

[28] ax" +ay' + a'' 2" + by' z' + b' z' x' + U'x'y' -^D' = 0, 

By comparing this equation with [8], [9] and [10], it wnll 
be seen that the cofficients a, a' , a" , b, b\ b" depend for their 
values upon <^, a', a'\ /^, /5', /?", y^ y', y" , and upon the co- 
efficients A, B, C, etc., of [i], § 440. The latter are known 
and thefo rmer are direction cosines. by § 407. 

Now let the axes 0"X'", 0"Y"', 0"Z'" be so chosen that 
in [28] 

[29] b = o, 

[30] b'--=o, 

[31] b''=o. 



CENTRAL QUADRICS 34^ 

Equations [29], [30] and [31], and§§ 411 and 412 give us 
nine equations from which the nine direction cosines may be 
found. 

When these are found and substituted into [28], that equa- 
tion will become by [29], [30] and [31] 

[32] ay^ + ay^ + ^v^ + i;>'^o. 

The accents over the variables being no longer needed, this 
equation may be written 

[33] «^^+«y + «v + z:)'=o, 

\ n 

[34] or ~";^-^ ~7/-^"~^^^' 

which is the equation of the quadric referred to the axes 
0"X"^ 0"Y"', 0"Z'". 

Now since the nine direction cosines have been found, the 
coefficients of [34] will depend only upon the coefiScients A, 
B, C, A\ etc., of [i], § 440, and will be positive or negative 
according as these latter are positive or negative. 

Let us take the positive value of — and let \/-7c, = — . 

[35] Then ^ = -,. 

Now since the a of [34] may be either positive or nega- 
tive, the coefficient ^ may be either positive or negative. 

Hence by [35] we get 

[36] -^^^ = ±-,. 



B' a 



Similarly w^e may get 

[37] -77-^=^ Tr 

a" z^ 

[38] and — -^, z'' = ± — ,. 

Hence by [36], [37] and [38], equation [34] may be written 

2 O .) 

±^±^±1^ = 1 

a/ b^ c;' ' 



346 



ANALYTIC GEOMETRY 






= I, 



or dropping subscripts, 

[A] ± ^ 

a' 

which is the equation of the central quadric. q. k. d. 

448. Equation [A], § 447, the equation of the central quadric, 
may take four different forms : 

ist. 



a' "^ b' "^V - '• 



2nd. 



d'^ b' T~^ 



A ^ 



4th. — 



X' 



a' 






+^ = . 



449. The Ellipsoid.— If every section of a quadric made by 
a plane parallel to one of the coordinate planes is an ellipse, 
the quadric is called an ellipsoid. 

PROPOSITION n 

450. The equatio7i of the ellipsoid is 



X' 



a' 



y 






in whicha, band c represent the semi-principal axes of the ellipsoid. 




Fig-. 1^6 



CENTRAL QUADRICS 347 

To determine the quadric represented by 

which is the first form of [A], § 447, we must determine the 
sections formed on this quadric by planes parallel to the coor- 
dinate planes. 

IvCt ABD'B'DA' be the quadric represented by [i] . 

I^et MN be a plane || to the XY plane. 

The equation of the plane MN is 

[2] z ^= p. by § 420. 

To determine the section formed on the quadric by the 
plane MN, we must treat [i] and [2] as simultaneous and 
solve them. by § 436. 

There will be three cases 

ist. when /><<:. 

2nd. p = c. 

3rd. / > ^. 

FIRST CASE 

P<c. 

Substituting the value of z found in [2] into [i] , we get 

[3] ^ + 7^+7^-'- 

[4] Hence _^ + -^==i_-. 

P' P' 

Since in this case /> < c, then ^ < i, hence i — ^ is 

c c 

positive. 

[5l Let ^-^ = ^'• 

Then by [4] and [5] w^e get 

[6] ^ + ^=?- 

^7] Hence -|-^^ + ^,= i, 



348 ANAL YTIC GEO ME TR Y 

which is the equation of the section EGE'H. by § 436. 

Now [7] is the equation of an ellipse whose semi-axes are 
aq and bq. by § 11 1. 

[A] Hence KGE'H is an ellipse whose semi-axes are aq 
and bq. 



By [5] we see 


that 


[s: 


q<i. 


I9] Hence 


aq << a, 


^10] and 


bq < b. 




SECOND CASE 




P=^c. 


Since in this case p = c, then ^ = 


]ii' 


q =0, 


and hence [6] becomes 


[12] 





I, hence by [5] we get 



[13] b^x- -\- a^yr = o. 

Since in [13] «V and b'^y' are necessarily positive, this 
equation can only be satisfied when ^ = o and jj/ = o, that is, 
the ellipse KGE'H becomes a point. 

[B] Therefore when p ^ c, the plane MN is tangent to 
the quadric. 

THIRD CASE 
P> c. 

Since in this case/ > c, then^ > i, hence by [5] ^^ will 

be negative and [6] becomes 

2 2 

[14] ^+^ = -?^ 

o 2 

V~ X 

[15] hence ■l^ = — —^—q\ 



CENTRAL QUADRICS 349 

Now from [15] we see that for every value of x,j/is 
imaginary. 

[C] Hence when /> > <: the plane MN cannot cut the 
quadric. 

Therefore from [A], [B] and [C] it follows that every sec- 
tion of the quadric represented by [i], made by a plane par- 
allel to the XY plane is an ellipse. 

Similarly it may be shown that every section of this quadric, 
made by a plane parallel to either of the other coordinate 
planes, is an ellipse. 

Therefore the quadric represented by [i] is an ellipsoid, 

by § 449. 

451. The Principal Axes. — 

When/^o, [5] becomes 

I = ?^ 

and [6] becomes 

x^ v^ 

[16] ^+i = '- 

which is the equation of the ellipse AD'A'D. 

By [9] and [10] we see that the axes of this ellipse are 
greater than those of any of the other ellipses made by planes 
parallel to the XY plane. 

Similarly we may show that the equation of the ellipse 

BD'B'D is 

2 2 

[17] ^ + ^ = 1- 

and that its axes d and c are greater than those of any of the 
other ellipses like JKL made by planes parallel to the ZY 
plane. 

Hence a, h and c are called the principal axes of the quadric. 

452. The Ellipsoid of Revolution. — The surface gener- 
ated by revolving an ellipse about either of its axes is called 
an ellipsoid of revolution. 



350 



ANALYTIC GEOMETRY 



453. Co7'ollary. — If any two of the axes of an ellipsoid be 
equal to each othei\ it will be an ellipsoid of revolutioji. 

For if in [i] b and c be equal to each other, then every 
section of the ellipsoid made by a plane |1 to the ZY plane 
will be a circle, and the ellipsoid in Fig. 159 may be gener- 
ated by revolving the ellipse AB'A'B about the axis AA'. 

If a > z^* = r, the ellipsoid is prolate. 
'^ a <ib = c, '' " " oblate. 

'' a = b = c, '' " "a sphere. 

454. The Hyperboloid of One Nappe. — If a quadric be 
continuous and every section of it made by a plane H to one 
of the coordinate planes be an ellipse and every section of it 
made by a plane || to either of the other coordinate planes be 
an hyperbola, the quadric is called an hyperboloid of one nappe. 



PROPOSITION III 

455. The equation of the hyperboloid of one nappe is 



X' 



"^ b' 



a c 

To determine the quadric represented by 



[i] 



2 2" 

X _, y z' 



which is the second form of [A] , § 448, we must determine 
the sections formed on it by planes |1 to the coordinate planes. 

Let ABA'B'DD' be the quadric represented by [i] . 
Let MN be a plane || to the XY plane. 
The equation of MN is 

[2] z =P' by § 420. 

Substituting this value of 2 into [i], we get 

X' 



[3] 



a 






which is the equation of the section HHE'G. 



by § 436. 



CENTRAL OUADRICS 



351 




Fig;. i5o 



I P 2 

q' > I, 



X' ^ y 



[4] Now let 

[5] Then 
and [3] becomes 

1^^] iaqY ' {bgY 

which is the equation of an ellipse w^hose semi- axes are aq 
and bq. by § 1 1 1 . 

If the plane MN be passed through the center C then in [2] 

[7] P = o, 

[8] and ^ = !> 

and [6] becomes 



'352 . ANALYTIC GEOMETRY 

2 '> 

r 1 X \ y 

[9] ^ +"7^ = ^' 

a o 

which is the equation of the ellipse Jly, whose semi-axes 
are a and b. 

Now by [5] 

[10] aq ^ a. 

[11] bq>b. 

Hence by [6] and [9] we see that the size of the ellipse in- 
creases continually as the plane MN is moved farther and 
farther from the center C, and that the quadric is continuous. 

JL is called the ellipse of the gorge. 

Let M'N' be a plane || to the ZY plane. Its equation will be 
[12] X ^ p. by § 420. 

Substituting the value of x found in [12] into [i] , we get 

[13] -^-^ = 1-^. 

which is the equation of the hyperbola ORSTUV. by § 204. 

Similarl}^ it may be proved that the equation of every sec- 
tion of the quadric formed by a plane || to the ZX plane is 

r 1 X z p 

[14] -^ r = I i-> 

^ -■ a c c 

which is also the equation of an hyperbola. 

Therefore by [6j, [13] and [14] we see that the quadric 
represented by [i] is an hyperboloid of one nappe, § 454. 

Q. K. D. 

456. The Hyperboloid of Revolution of One Nappe. — 

The surface generated by revolving an hyperbola about its 
conjugate axis is called an hyperboloid of revolution of one 
nappe. 

457. Corollary. — If the two axes a and b of the hyperboloid of 
one nappe are equal to each other it is an hyperboloid of rev- 
olution of one nappe. 



CENTRAL QUADRICS 



353 



458, The Hyperboloid of Two Nappes. — If a quadric is 
discontinuous and every section of it made by a plane || to one 
of the coordinate planes is an ellipse, and every section of it 
made by a plane || to either of the other coordinate planes is 
an hyperbola, the quadric is called an hyperboloid of two 
nappes. 



PROPOSITION IV 
459- 1^^^^ equation of a7i hyperboloid of two nappes is 

X' 

a' 



-2 ^,2 -2 




Fia:. 161 
The third form of equation [A] § 448 is 



[I] 



X' 

a 



r 






354 ANALYTIC GEOMETRY 

Dividing this equation through by — j, we get 

2 2 2 

Let ABA'B'DK be the quadric represented by [2]. 
Let MN be a plane || to the XY plane. 

Then, as in the preceding proposition, the equation of the 
section formed on the quadric by the plane MN is 



[3! 


a' 


+ 


i;=-- 


+ - 


[4] 


Let 




■+$= 


-q' 


then [3^ 


becomes 








[5] 




- 


^' 4 f _ 
a' "^ b' ~~ 


--q\ 


which is 


the equation 


of 


an ellipse. 




As in 


§ 450 there may 


be three cases 




ist. 




P <c. 






2nd. 




P = c. 






3rd. 




p> c. 





by § III. 



FIRST CASE 
p <C. 

In this case since /> < c, [4] shows that q^ must be negative 
and [3] becomes 



>; 




—q' 


[7] Hence 


i—r- 


x' 
a' 



[7] shows that for every real value of x, y is imaginary, 
that is, the ellipse is imaginary. 

Therefore the plane MN does not cut the quadric repre- 
sented by .[2] so long as /> < <:. 

Hence the quadric represented by [2] is discontinuous. 



CENTRAL QUADRICS 355 

SECOND CASE 

In this case [3] becomes 

[8] i?+i=°' 

and, as in §450, [13], it can be shown that the section be- 
comes a point. 

Therefore the plane is tangent to the quadric when/ = c, 

THIRD CASE 

P>c. 

In this case since/ > c, [4] shows that q^ must be positive 
and [3] becomes 

[9] ^ + 7^ = ?^ 

2 2 

[10] hence _^^ + ^^=i, 

which is the equation of an ellipse whose semi-axes are aq 
and bq. by § 1 11. 

Now by [4] we see that after p becomes equal to c, q in- 
creases continually as the plane MN is moved farther and 
farther from O. Hence the semi-axes <2^ and bq must also in- 
crease. 

Therefore the ellipse HHE'G gets larger and larger as MN 
is moved farther and farther from the origin. 

As in the preceding proposition, it may be shown that every 
section of the quadric made by any plane M'N' i| to the ZY 
plane is an hyperbola, and that every section of it made by 
any plane || to the ZX plane is also an hyperbola. 

Since the quadric represented by [2] is discontinuous and 
every section of it made by a plane || to the XY plane is an 
ellipse, and every section of it made by a plane |i to either of 
the other coordinate planes is an hyperbola, therefore the 
quadric must be an hyperboloid of two nappes. by § 458. 

O. E. D. 



356 



ANALYTIC GEOMETRY 



460. The Hyperboloid of Revolution of Two Nappes. — 

The surface generated by revolving an hyperbola about its 
transverse axis is called an hyperboloid of revolution of two 
nappes. 

461. Corollary . — If the two imaginary axes a and b of the 
hyperboloid of two nappes are equal to each other, it is an hyper- 
boloid of revolutiofi of two ?iappes. 

462. Scholium. — The fourth form of [A], §448, does not 
represent a real surface. For dividing it through by — i, 
we get 



[I] 



,2 ~r i2 



+ -:x= — I- 



The equation of au}^ plane H to the XY plane is 

[2] z—p. 

Substituting this value of z into [i] we get 



by §420. 



[3] 



^2 l~ i2 ^ Ji. 1 

a c 



which is the equation of the section formed on the surface rep- 
resented by [i] by any plane || to the XY plane. by § 436. 

From [3] we get 

Since every term in the second member of this equation is 
necessarily negative, jk must be imaginary for every real value 
of ;i; in it. 

That is, no plane || to the XY plane and at a finite distance 
from the origin can cut the surface represented by [i]. 

Similarly it may be showm that no plane |1 to either of the 
other coordinate planes can cut the surface represented by [i]. 

Therefore there can be no real surface represented by [i]. 



CHAPTER X 

Non-Central Quadrics 

463. On page 343, §444, we have shown that the quadric 
will be non-central if 



[i] 



A 


B" 


B' 


B" 


A' 


B 


B' 


B 


A" 



^= o. 



In the more extensive treatises on this subject it is shown 
that when this determinant is o, [i] § 440 can be reduced to 
the form 



[B] 



^y + ^v + 2C^ = o. 



Equation [B] may take two different forms according as 
the signs of A' and ^" are alike or unlike. 

ist. Ay + ^ V — 2Cx. 
2nd. Ay — A^'s' — 2Cx. 



464. The Elliptic Paraboloid. — If every section of a non- 
central quadric made by a plane || to one of the coordinate 
planes is an ellipse, and the sections formed on it by planes || 
to the other two coordinate planes are parabolas, then the 
quadric is called an elliptic paraboloid. 



358 



ANAL YTIC GEO ME TR Y 



PROPOSITION I 
465. The equatio7i of an elliptic paraboloid is 




Fig. 162 

The first form of [B] , § 463, is ' 

Let MN be a plane 1| to the XY plane. Its equation is 
[2] ^ — p. by §420. 

The equation of the section CBP, made b)^ the plane MN 
on the quadric represented by [i] , is 

[3] Ay = 2Cx — A"p\ by §436. 

which is the equation of a parabola. by § 287. 

Hence every section of the quadric made by a plane 1| to the 
XY plane is a parabola. 

The equation of a section made by a plane || to the ZX 
plane is 

[4] A"^' = 2Cx — Ap\ by § 436. 

which is also the equation of a parabola. by § 287. 



NON-CENTRAL QUADRICS. 359 

Hence every section of the quadric made by a plane 1| to the 
ZX plane is a parabola. 

The equation of the section AEGH, made by the plane 
M'N' II to the ZY plane is 

[5] ^y + ^v = 2rA by §436. 

which is the equation of an ellipse. by § iii. 

Since every section of the quadric represented by [i] made 
by a plane i| to the ZY plane is an ellipse, and every section 
of it made by a plane |1 to either of the other coordinate planes 
is a parabola the quadric is an elliptic paraboloid. by § 464. 

466. The Hyperbolic Paraboloid. — If every section of a 
non-central quadric made by a plane || to one of the coordi- 
nate planes is an hyperbola, and every section of it made by 
planes || to the other coordinate planes is a parabola, the 
quadric is called an hyperbolic pa^^aboloid. 

467. The equation of an hyperbolic paraboloid is 

Ay — A"2' z= 2Cx. 

The second form of [B] , § 463, is 

[i] A'y' — ^'V — 2Cx. 

Let MN be any plane |1 to the ZY plane. Its equation is 

[2] X :^ p. by § 420. 

The equation of the section ADEA'GH, made by the plane 
MN on the quadric represented by [i] , is 

[3] AY — A"2' = 2Cp, by § 436. 

which is the equation of an h^^perbola. by § 204. 

Hence every section of the quadric represented by [i], 
made by a plane || to the ZY plane, is an hyperbola. 



360 



ANAL YTIC GEO ME TR Y 




Fig. 163 

Similarly every section of the quadric made by a plane |I to 
the ZX plane, is also an hyperbola. 

The equation of the section RST, made by any plane || to 
the XY plane, is 

A'y^ = 2Cx + A''p\ by § 436. 

which is the equation of a parabola. by § 287. 

Hence every section of the quadric represented by [i], made 
by a plane || to the XY plane, is a parabola. 

Since every section of the quadric represented by [i] , made 
by a plane || to the XY plane, is a parabola, and every section 
of it made by a plane || to either of the other coordinate 
planes, is an hyperbola, the quadric is an hyperbolic paraboloid. 

by § 466. 



APPENDIX 



PROPOSITIONS REFERRED TO IN THE TEXT 



GEOMETRY 

1. Things which are equal to the same thing are equal to each other. 

2. At any given point in a given straight line one perpendicular and 
only one can be erected. 

3. If two adjacent angles have their exterior sides in a straight line, 
these angles are supplements of each other. 

4. If one straight line intersects another straight line, the vertical 
angles are equal. 

5. From a point without a straight line one perpendicular, and only 
one, can be drawn to this line. 

6. If a straight line is perpendicular to one of two parallel lines, it 
is perpendicular to the other also. 

7. If two parallel straight lines are cut by a third straight line, the 
alternate interior angles are equal. 

8. If two parallel straight lines are cut by a third straight line, the 
exterior interior angles are equal. 

9. When two straight lines are cut by a third straight line, so as to 
make the exterior interior angles equal, these two straight lines are 
parallel. 

10. Two straight lines which are parallel to a third straight line, are 
parallel to each other. 

11. Two angles whose sides are parallel, each to each, are either 
equal or supplementary. 

12. Two angles whose sides are perpendicular, each to each, are 
either equal or supplementary. 

13. The exterior angle of a triangle is equal to the sum of the two 
opposite interior angles. 

14. Two right angles are equal if the hypotenuse and an acute angle 
of one are equal respectively to the hypotenuse and an acute angle of 
the other. 

15. Two right triangles are equal if their legs are equal, each to 
each. 



362 APPENDIX 

i6. In an isosceles triangle the sides opposite the equal angles are 
equal. 

17. Parallel lines comprehended between parallel lines are equal, 

18. A circle is a portion of a plane bounded by a curved line called 
a circumference, all points of which are equally distant from a point 
within called the center. 

19. The limit of a variable is 

(i) a constant, 

(2) towards which the variable continually approaches, 

(3) and from which it may be made to differ by a quantity 

which is less than a given positive quantity however 
small this latter may be made. 

20. If two variables are constantly equal and each approaches a 
limit, the limits are equal. 

21. In every proportion the product of the extremes is equal to the 
product of the means. 

22. In a series of equal ratios, the sum of the antecedents is to the 
sum of the consequents as any antecedent is to its consequent. 

23. If a line be drawn through two sides of a triangle parallel to the 
third side, it divides those sides proportionally. 

24. If a straight line divides two sides of a triangle proportionally, 
it is parallel to the third side, 

25. If two triangles have their sides respectively parallel, or respect- 
ively perpendicular, they are similar. 

26. The sum of the squares of the two legs of a right triangle is equal 
to the square of the hypotenuse. 

27. The square of either leg of a right triangle is equal to the differ- 
ence of the squares of the hypotenuse and the other leg, 

28. The area of a rectangle is equal to the product of its base and 
altitude. 

29. The ratio of the circumference of a circle to its diameter is con- 
stant. C = 27rR. 

30. The area of a circle equals -k times the square of its radius. 

31. Similar polygons are polygons having their homologous sides 
proportional and their homologous angles equal, 

32. The square on any line is four times the square on half the line. 

33. A straight line is perpendicular to a plane if it is perpendicular 
to every straight line of the plane drawn through its foot ; that is, 
through the point where it meets the plane. 

34. The projection of a point on a plane is the foot of the perpen- 
dicular from the point to the plane. 



APPENDIX 363 

35. Two straight lines perpendicular to the same plane are parallel. 

36. If one of two parallel lines is perpendicular to a plane, the 
other is also perpendicular to the plane. 

37. If tW'O straight lines are parallel to a third straight line they are 
parallel to each other. 

38. Two planes perpendicular to the same straight line are parallel. 

39. The intersections of two parallel planes with a third plane are 
parallel lines. 

40. Parallel lines included between parallel planes are equal. 

41. If two angles not in the same plane have their sides respectively- 
parallel and lying in the same direction, they are equal, and their 
planes are parallel. 

42. If two planes are perpendicular to each other, a perpendicular 
to one of them at any point of their intersection will lie in the other. 

43. If a straight line is perpendicular to a plane, every plane passed 
through the line is perpendicular to the first plane. 

44. If two intersecting planes are each perpendicular to a third 
plane, their intersection is also perpendicular to that plane. 

45. Through a given straight line not perpendicular to a plane, one 
plane, and only one, can be passed perpendicular to the given plane. 

46. Two straight lines in the same plane perpendicular to the same 
line are parallel. 

47. In a parallelogram the opposite sides are equal and the opposite 
angles are equal. 

48. If two sides of a quadrilateral are equal and parallel, then the 
other two sides are equal and parallel, and the figure is a parallelo- 
gram. 

49. A straight line is inscribed in a circle if it is a chord. 

50. Two triangles are similar if two angles of the one are respect- 
ively equal to two angles of the other. 

51. Two right triangles are similar if an acute angle of the one is 
equal to an acute angle of the other. 

53. At a given point in a straight line one plane perpendicular to 
the line can be drawn, and only one. 

54. See 51. 

55. An angle inscribed in a semicircle is a right angle. 

56. If the product of any two factors be equal to the product of any 
other two factors, we may take the factors of either product as the 
means of a proportion if we take the factors of the other product as 
the extremes. 



364 APPENDIX 

57. In the same circle or equal circles, chords equally distant from 
the center are equal. 

58. Bqual chords subtend equal arcs. 

59. A radius perpendicular to a chord bisects the chord and also the 
arc which it subtends. 

60. The perpendicular from any point in the circumference to the 
diameter of a circle is a mean proportional between the segments of 
the diameter. 

61. The area of a trapezoid is equal to one-half the sum of the 
parallel sides multiplied by the altitude. 



TRIGONOMETRY. 



a opposite leg 

1. sm A = = ^^ , ^. 

c hypotenuse 

b adiacent leg 

2. cos A = = — ^. 

c hypotenuse 

a opposite leg 

b adjacent leg 

b adjacent leg 



a opposite leg 

b adjacent leg 



4. cot A 



a opposite leg' 

5. sin^ A 4- cos^ A =^ \. 

sin A 

6. tan A = -.• 

cos A 

7. sin A X cosec A =^ \. 

8. cos A X sec ^ — i. 

9. tan A X cot A =^ \. 

10. I + cot^ A = cosec^ A. 

11. sin ( — A) = —sinA, tan ( — A) = — tan A. 

12. cos ( — A)=cosA, cot ( — A) :;= — cot ^. 

13. sin (A — B) = sin A cos B — cos A sin B. 

14. The sides of a triangle are proportional to the sines of the op- 
posite angles. 

15. The area of a parallelogram is equal to the product of any two 
adjacent sides by the sine of the included angle. 

16. In quadrant 2 the sine and cosecant only are positive. 

. ^ ^, tan A — tan B 

17. tan {A — B)= — j— ^- ^. 

^ I + tan A tan B 



APPENDIX 365 



sin 



M - ± J ' ~ ^^' ^ ; cosi^ = ±J^ 



-f- cos A 
2 



0°. 90°. 180°. 270°. 360°. 

cosine i ±0 — i ±0 i 

■'■^' tangent ±0 ±00 ±0 ±00 ±0 

20. sin A = cos B, cos A = sin B, when A -\- B ^=^ 90. 

21. cos (90 + A') ■= — sin A. 

22. sin (180 — ^) = sin ^. 

23. cos (180 — A) = — cos y^. 

24. tan (180 — A) = — tan A. 

25. The square of any side of a triangle is equal to the sum of the 
squares of the other two sides, diminished by twice their product into 
the cosine of the included angle. 

26. tan 45'^ = I. 

27. A radian is the angle at the center of a circle subtended by an 
arc whose length is equal to the radius of the circle. It is equal to 57°. 3 

28. The number of radians in a given angle is equal to its arc divided 
by the radius of the circle. 

29. vers A = 1 — cos A. 

_ r — -y T — y 

30. cos ' is read thus : the arc whose cosine is . 

r r 

31. The area of a triangle is equal to half the product of two adjacent 
sides by the sine of the included angle. 



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Notes on Qualitative Analysis. By W. P. Mason, Professor 
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Mit einer grossen Sorgfalt und Selbstandigkeit hat der Verf. alles 
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Wir haben es somit mit einer wertvollen und beachtenswerten Arbeit 
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Lubricating Oils, Fats, and Greases. Their Origin, Prepa- 
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Methods for the Analysis of Ores, Pig Iron and Steel in 
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The Iron Blast Furnace. By G. H. Meeker, formerly 
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